1 The IMO Compendium - imomath.com · 1 Djuki ć · Jankovi ć Mati ć · Petrovi ć Problem Books in Mathematics Dušan Djukić · Vladimir Janković · Ivan Matić · Nikola Petrović

1

Djuki! · Jankovi! M

ati! · Petrovi!

Problem Books in Mathematics

Du"an Djuki! · Vladimir Jankovi! · Ivan Mati! · Nikola Petrovi!

The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009, Second Edition


The IMO Compendium

Du"an Djuki!Vladimir Jankovi!Ivan Mati!Nikola Petrovi!

Mathematics

ABCDPBM

The IMO Com

pendium

A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009

Second Edition

! e IMO Compendium is the ultimate collection of challenging high school level mathemat-ics problems. It is an invaluable resource, not only for students preparing for competitions, but for anyone who loves and appreciates math. Training for mathematical olympiads is enjoyed by talented students throughout the world. Olympiads have become one of the primary ways to recognize and develop talented youth with a potential to excel in areas that require abstract thinking. Although the problems appearing at IMO do not involve advanced mathematics, they must be di! cult and their solutions must arise from creative and clever insights rather than tedious calculations.

In preparation for the distinguished International Mathematical Olympiad (IMO) competi-tion, each participating country selects the top six high school students every year through a series of national olympiads. " ese students are invited to participate in the IMO, usually held in July. " e IMO is a two-day contest where each day competitors are given three problems which they work on independently. " e IMO host country appoints a special committee to which each country submits up to six problems. From this composite “longlist” of prob-lems, a “shortlist” of #$-%& problems is created. " e jury, consisting of one professor from each country, makes the ' nal selection from the shortlist a few days before the IMO begins.

" e IMO has sparked a burst of creativity among enthusiasts to create new and interest-ing mathematics problems. It can be safely said that the IMO and shortlisted problems are among the well-cra( ed problems created in a given year. " is book attempts to gather all of these problems with their solutions. In addition, the book contains all the available longlist problems, for a total of more than #&&& problems.

From the reviews of the ' rst edition:

"" e International Mathematical Olympiad, or IMO is the premier international competi-tion for talented high school mathematics students. … " is book collects statements and solutions of all of the problems ever set in the IMO, together with many problems proposed for the contest. … serves as a vast repository of problems at the Olympiad level, useful both to students … and to faculty looking for hard elementary problems. No library will want to be without a copy, nor will anyone involved in mathematics competitions …"

— (Fernando Q. Gouvêa, MathDL, March, #&&))

2nd Ed.9 781441 998538

ISBN 978-1-4419-9853-8

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The International Mathematical Olympiad . . . . . . . . . . . . . . . . . . . . . . 11.2 The IMO Compendium. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Basic Concepts and Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.1.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.2 Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.3 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.4 Groups and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.1 Triangle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3.2 Vectors in Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.3 Barycenters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3.4 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3.5 Circle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3.6 Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3.7 Geometric Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.8 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.9 Formulas in Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.4 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4.1 Divisibility and Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4.2 Exponential Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.4.3 Quadratic Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . 212.4.4 Farey Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.5 Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5.1 Counting of Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5.2 Graph Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

X Contents

3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.1 IMO 1959 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 IMO 1960 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29







3.8.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.8.2 Some Longlisted Problems 1959–1966 . . . . . . . . . . . . . . . . . . 35

3.9 IMO 1967 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.9.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.9.2 Longlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.10 IMO 1968 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.10.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.10.2 Shortlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49


3.12 IMO 1970 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.12.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.12.2 Longlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.12.3 Shortlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68





Contents XI

3.16.3 Shortlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.17 IMO 1975 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

3.17.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 973.17.2 Shortlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97











3.28 IMO 1987 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1923.28.1 Contest Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

XII Contents

3.28.2 Longlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1923.28.3 Shortlisted Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200














Contents XIII










4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3474.1 Contest Problems 1959 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3474.2 Contest Problems 1960 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3494.3 Contest Problems 1961 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3514.4 Contest Problems 1962 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3534.5 Contest Problems 1963 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3544.6 Contest Problems 1964 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3554.7 Contest Problems 1965 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3574.8 Contest Problems 1966 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3594.9 Longlisted Problems 1967 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3614.10 Shortlisted Problems 1968 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3744.11 Contest Problems 1969 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3804.12 Shortlisted Problems 1970 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3834.13 Shortlisted Problems 1971 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3894.14 Shortlisted Problems 1972 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3964.15 Shortlisted Problems 1973 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4014.16 Shortlisted Problems 1974 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407

XIV Contents

4.17 Shortlisted Problems 1975 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4134.18 Shortlisted Problems 1976 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4184.19 Longlisted Problems 1977 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4224.20 Shortlisted Problems 1978 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4374.21 Shortlisted Problems 1979 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4454.22 Shortlisted Problems 1981 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4534.23 Shortlisted Problems 1982 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4614.24 Shortlisted Problems 1983 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4674.25 Shortlisted Problems 1984 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4764.26 Shortlisted Problems 1985 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4834.27 Shortlisted Problems 1986 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4914.28 Shortlisted Problems 1987 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4984.29 Shortlisted Problems 1988 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5084.30 Shortlisted Problems 1989 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5234.31 Shortlisted Problems 1990 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5374.32 Shortlisted Problems 1991 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5504.33 Shortlisted Problems 1992 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5634.34 Shortlisted Problems 1993 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5734.35 Shortlisted Problems 1994 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5854.36 Shortlisted Problems 1995 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5934.37 Shortlisted Problems 1996 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6064.38 Shortlisted Problems 1997 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6224.39 Shortlisted Problems 1998 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6364.40 Shortlisted Problems 1999 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6504.41 Shortlisted Problems 2000 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6644.42 Shortlisted Problems 2001 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6774.43 Shortlisted Problems 2002 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6914.44 Shortlisted Problems 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7024.45 Shortlisted Problems 2004 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7154.46 Shortlisted Problems 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7304.47 Shortlisted Problems 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7424.48 Shortlisted Problems 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7544.49 Shortlisted Problems 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7654.50 Shortlisted Problems 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777

A Notation and Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791A.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791A.2 Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792

B Codes of the Countries of Origin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795

C Authors of Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 797

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805

322 3 Problems

3.46 The Forty-Sixth IMOMérida, Mexico, July 8–19, 2005

3.46.1 Contest Problems

First Day (July 13)

1. Six points are chosen on the sides of an equilateral triangle ABC: A1,A2 on BC;B1,B2 on CA; C1,C2 on AB. These points are vertices of a convex hexagonA1A2B1B2C1C2 with equal side lengths. Prove that the lines A1B2, B1C2 andC1A2 are concurrent.

2. Let a1,a2, . . . be a sequence of integers with infinitely many positive terms andinfinitely many negative terms. Suppose that for each positive integer n, the num-bers a1,a2, . . . ,an leave n different remainders on division by n. Prove that eachinteger occurs exactly once in the sequence.

3. Let x, y, and z be positive real numbers such that xyz! 1. Prove that

x5" x2

x5+ y2+ z2+

y5" y2

y5+ z2+ x2+

z5" z2

z5+ x2+ y2! 0.

Second Day (July 14)

4. Consider the sequence a1,a2, . . . defined by

an = 2n+3n+6n"1 (n= 1,2, . . .).

Determine all positive integers that are relatively prime to every term of thesequence.

5. Let ABCD be a given convex quadrilateral with sides BC and AD equal in lengthand not parallel. Let E and F be interior points of the sides BC and AD respec-tively such that BE =DF . The lines AC and BD meet at P; the lines BD and EFmeet at Q; the lines EF and AC meet at R. Consider all the triangles PQR as Eand F vary. Show that the circumcircles of these triangles have a common pointother than P.

6. In a mathematical competition, six problems were posed to the contestants. Eachpair of problems was solved by more than 2/5 of the contestants. Nobody solvedall six problems. Show that there were at least two contestants who each solvedexactly five problems.

3.46.2 Shortlisted Problems

1. A1 (ROU) Find all monic polynomials p(x) with integer coefficients of degreetwo for which there exists a polynomial q(x) with integer coefficients such thatp(x)q(x) is a polynomial having all coefficients±1.

3.46 IMO 2005 323

2. A2 (BGR) Let R+ denote the set of positive real numbers. Determine all func-tions f :R+ # R+ such that

f (x) f (y) = 2 f (x+ y f (x))

for all positive real numbers x and y.

3. A3 (CZE) Four real numbers p,q,r,s satisfy

p+q+ r+ s= 9 and p2+q2+ r2+ s2 = 21.

Prove that ab" cd ! 2 holds for some permutation (a,b,c,d) of (p,q,r,s).

4. A4 (IND) Find all functions f : R # R satisfying the equation

f (x+ y)+ f (x) f (y) = f (xy)+2xy+1

for all real x and y.

5. A5 (KOR)IMO3 Let x,y and z be positive real numbers such that xyz ! 1. Provethat

x5" x2

x5+ y2+ z2+

y5" y2

y5+ z2+ x2+

z5" z2

z5+ x2+ y2! 0.

6. C1 (AUS) A house has an even number of lamps distributed among its roomsin such a way that there are at least three lamps in every room. Each lamp sharesa switch with exactly one other lamp, not necessarily from the same room. Eachchange in the switch shared by two lamps changes their states simultaneously.Prove that for every initial state of the lamps there exists a sequence of changesin some of the switches at the end of which each room contains lamps that areon as well as lamps that are off.

7. C2 (IRN) Let k be a fixed positive integer. A company has a special methodto sell sombreros. Each customer can convince two persons to buy a sombreroafter he/she buys one; convincing someone already convinced does not count.Each of these new customers can convince two others and so on. If each of thetwo customers convinced by someone makes at least k persons buy sombreros(directly or indirectly), then that someone wins a free instructional video. Provethat if n persons bought sombreros, then at most n/(k+2) of them got videos.

8. C3 (IRN) In anm$n rectangular board ofmn unit squares, adjacent squares areones with a common edge, and a path is a sequence of squares in which any twoconsecutive squares are adjacent. Each square of the board can be colored blackor white. Let N denote the number of colorings of the board such that there existsat least one black path from the left edge of the board to its right edge, and letMdenote the number of colorings in which there exist at least two nonintersectingblack paths from the left edge to the right edge. Prove that N2 ! 2mnM.

9. C4 (COL) Let n ! 3 be a given positive integer. We wish to label each sideand each diagonal of a regular n-gon P1 . . .Pn with a positive integer less than orequal to r so that:

324 3 Problems

(i) every integer between 1 and r occurs as a label;(ii) in each triangle PiPjPk two of the labels are equal and greater than the third.Given these conditions:(a) Determine the largest positive integer r for which this can be done.(b) For that value of r, how many such labelings are there?

10. C5 (SCG) There are n markers, each with one side white and the other sideblack, aligned in a row so that their white sides are up. In each step, if possible,we choose a marker with the white side up (but not one of the outermost mark-ers), remove it, and reverse the closest marker to the left and the closest markerto the right of it. Prove that one can achieve the state with only two markersremaining if and only if n"1 is not divisible by 3.

11. C6 (ROU)IMO6 In a mathematical competition, six problems were posed to thecontestants. Each pair of problems was solved by more than 2/5 of the contes-tants. Nobody solved all six problems. Show that there were at least two contes-tants who each solved exactly five problems.

12. C7 (USA) Let n! 1 be a given integer, and let a1, . . . ,an be a sequence of inte-gers such that n divides the sum a1+ · · ·+an. Show that there exist permutations! and " of 1,2, . . . ,n such that !(i)+ "(i) % ai (mod n) for all i= 1, . . . ,n.

13. C8 (BGR) Let M be a convex n-gon, n ! 4. Some n" 3 of its diagonals arecolored green and some other n" 3 diagonals are colored red, so that no twodiagonals of the same color meet insideM. Find the maximum possible numberof intersection points of green and red diagonals insideM.

14. G1 (HEL) In a triangle ABC satisfying AB+BC = 3AC the incircle has centerI and touches the sides AB and BC at D and E , respectively. Let K and L be thesymmetric points ofD and E with respect to I. Prove that the quadrilateralACKLis cyclic.

15. G2 (ROU)IMO1 Six points are chosen on the sides of an equilateral triangle ABC:A1,A2 on BC; B1,B2 on CA;C1,C2 on AB. These points are vertices of a convexhexagon A1A2B1B2C1C2 with equal side lengths. Prove that the lines A1B2, B1C2andC1A2 are concurrent.

16. G3 (UKR) Let ABCD be a parallelogram. A variable line l passing through thepoint A intersects the rays BC and DC at points X and Y , respectively. Let K andL be the centers of the excircles of triangles ABX and ADY , touching the sidesBX and DY , respectively. Prove that the size of angle KCL does not depend onthe choice of the line l.

17. G4 (POL)IMO5 Let ABCD be a given convex quadrilateral with sides BC and ADequal in length and not parallel. Let E and F be interior points of the sides BCand AD respectively such that BE = DF . The lines AC and BD meet at P; thelines BD and EF meet at Q; the lines EF and AC meet at R. Consider all thetriangles PQR as E and F vary. Show that the circumcircles of these triangleshave a common point other than P.

3.46 IMO 2005 325

18. G5 (ROU) Let ABC be an acute-angled triangle with AB &= AC; let H be itsorthocenter and M the midpoint of BC. Points D on AB and E on AC are suchthat AE = AD and D,H,E are collinear. Prove that HM is orthogonal to thecommon chord of the circumcircles of triangles ABC and ADE .

19. G6 (RUS) The median AM of a triangle ABC intersects its incircle # at K andL. The lines through K and L parallel to BC intersect # again at X and Y . Thelines AX and AY intersect BC at P and Q. Prove that BP=CQ.

20. G7 (KOR) In an acute triangle ABC, let D, E , F , P, Q, R be the feet of perpen-diculars from A, B, C, A, B, C to BC, CA, AB, EF , FD, DE , respectively. Provethat p(ABC)p(PQR) ! p(DEF)2, where p(T ) denotes the perimeter of triangleT .

21. N1 (POL)IMO4 Consider the sequence a1,a2, . . . defined by

an = 2n+3n+6n"1 (n= 1,2, . . .).

Determine all positive integers that are relatively prime to every term of thesequence.

22. N2 (NLD)IMO2 Let a1,a2, . . . be a sequence of integers with infinitely manypositive terms and infinitely many negative terms. Suppose that for each positiveinteger n, the numbers a1,a2, . . . ,an leave n different remainders on division byn. Prove that each integer occurs exactly once in the sequence.

23. N3 (MNG) Let a, b, c, d, e, and f be positive integers. Suppose that the sumS= a+b+c+d+e+ f divides both abc+de f and ab+bc+ca"de"e f " f d.Prove that S is composite.

24. N4 (COL) Find all positive integers n> 1 for which there exists a unique integera with 0< a' n! such that an+1 is divisible by n!.

25. N5 (NLD) Denote by d(n) the number of divisors of the positive integer n. Apositive integer n is called highly divisible if d(n) > d(m) for all positive integersm< n. Two highly divisible integers m and n with m< n are called consecutiveif there exists no highly divisible integer s satisfying m< s< n.(a) Show that there are only finitely many pairs of consecutive highly divisible

integers of the form (a,b) with a | b.(b) Show that for every prime number p there exist infinitely many positive

highly divisible integers r such that pr is also highly divisible.

26. N6 (IRN) Let a and b be positive integers such that an + n divides bn + n forevery positive integer n. Show that a= b.

27. N7 (RUS) Let P(x) = anxn+an"1xn"1+ · · ·+a0, where a0, . . . ,an are integers,an > 0, n ! 2. Prove that there exists a positive integer m such that P(m!) is acomposite number.

326 3 Problems

3.47 The Forty-Seventh IMOLjubljana, Slovenia, July 6–18, 2006

3.47.1 Contest Problems

First Day (July 12)

1. Let ABC be a triangle with incenter I. A point P in the interior of the trianglesatisfies

!PBA+!PCA= !PBC+!PCB.

Show that AP! AI, and that equality holds if and only if P= I.

2. Let P be a regular 2006-gon. A diagonal of P is called good if its endpointsdivide the boundary of P into two parts, each composed of an odd number ofsides ofP . The sides ofP are also called good.SupposeP has been dissected into triangles by 2003 diagonals, no two of whichhave a common point in the interior ofP . Find the maximum number of isosce-les triangles having two good sides that could appear in such a configuration.

3. Determine the least real numberM such that the inequality!!ab(a2"b2)+bc(b2" c2)+ ca(c2"a2)

!! 'M(a2+b2+ c2)2

holds for all real numbers a, b, and c.

Second Day (July 13)

4. Determine all pairs (x,y) of integers such that

1+2x+22x+1 = y2.5. Let P(x) be a polynomial of degree n> 1 with integer coefficients and let k be apositive integer. Consider the polynomial

Q(x) = P(P(. . .P(P(x)) . . . )),

where P occurs k times. Prove that there are at most n integers t that satisfy theequality Q(t) = t.

6. Assign to each side b of a convex polygonP the maximum area of a triangle thathas b as a side and is contained in P . Show that the sum of the areas assignedto the sides ofP is at least twice the area ofP .

3.47.2 Shortlisted Problems

1. A1 (EST) A sequence of real numbers a0, a1, a2, . . . is defined by the formula

ai+1 = [ai] · {ai}, for i! 0;

here a0 is an arbitrary number, [ai] denotes the greatest integer not exceeding ai,and {ai} = ai" [ai]. Prove that ai = ai+2 for i sufficiently large.

3.47 IMO 2006 327

2. A2 (POL) The sequence of real numbers a0, a1, a2, . . . is defined recursivelyby a0 = "1 and

n

$k=0

an"kk+1

= 0, for n! 1.

Show that an > 0 for n! 1.

3. A3 (RUS) The sequence c0, c1, . . . , cn, . . . is defined by c0 = 1, c1 = 0, andcn+2 = cn+1+ cn for n ! 0. Consider the set S of ordered pairs (x,y) for whichthere is a finite set J of positive integers such that x = $ j(J c j, y = $ j(J c j"1.Prove that there exist real numbers %,& , and M with the following property: anordered pair of nonnegative integers (x,y) satisfies the inequalitym<%x+&y<M if and only if (x,y) ( S.Remark: A sum over the elements of the empty set is assumed to be 0.

4. A4 (SRB) Prove the inequality

$i< j

aia jai+a j

'n

2(a1+a2+ · · ·+an)$i< jaia j

for positive real numbers a1,a2, . . . ,an.

5. A5 (KOR) Let a, b, c be the sides of a triangle. Prove that)b+ c"a)

b+)c"

)a

+

)c+a"b

)c+

)a"

)b

+

)a+b" c

)a+

)b"

)c' 3.

6. A6 (IRL)IMO3 Determine the smallest numberM such that the inequality

|ab(a2"b2)+bc(b2" c2)+ ca(c2"a2)|'M(a2+b2+ c2)2

holds for all real numbers a, b, c

7. C1 (FRA) We have n ! 2 lamps L1, . . . ,Ln in a row, each of them being eitheron or off. Every second we simultaneously modify the state of each lamp asfollows: if the lamp Li and its neighbors (only one neighbor for i = 1 or i = n,two neighbors for other i) are in the same state, then Li is switched off; otherwise,Li is switched on.Initially all the lamps are off except the leftmost one which is on.(a) Prove that there are infinitely many integers n for which all the lamps will

eventually be off.(b) Prove that there are infinitely many integers n for which the lamps will

never be all off.

8. C2 (SRB)IMO2 A diagonal of a regular 2006-gon is called odd if its endpointsdivide the boundary into two parts, each composed of an odd number of sides.Sides are also regarded as odd diagonals. Suppose the 2006-gon has been dis-sected into triangles by 2003 nonintersecting diagonals. Find the maximum pos-sible number of isosceles triangles with two odd sides.

328 3 Problems

9. C3 (COL) Let S be a finite set of points in the plane such that no three of themare on a line. For each convex polygon P whose vertices are in S, let a(P) bethe number of vertices of P, and let b(P) be the number of points of S that areoutside P. Prove that for every real number x

$Pxa(P)(1" x)b(P) = 1,

where the sum is taken over all convex polygons with vertices in S.Remark. A line segment, a point, and the empty set are considered convex poly-gons of 2, 1, and 0 vertices respectively.

10. C4 (TWN) A cake has the form of an n$n square composed of n2 unit squares.Strawberries lie on some of the unit squares so that each row and each columncontains exactly one strawberry; call this arrangementA .LetB be another such arrangement. Suppose that every grid rectangle with onevertex at the top left corner of the cake contains no fewer strawberries of arrange-mentB than of arrangementA . Prove that arrangementB can be obtained fromA by performing a number of switches, defined as follows:A switch consists in selecting a grid rectangle with only two strawberries, situ-ated at its top right corner and bottom left corner, and moving these two straw-berries to the other two corners of that rectangle.

11. C5 (ARG) An (n,k)-tournament is a contest with n players held in k roundssuch that:(i) Each player plays in each round, and every two players meet at most once.(ii) If player A meets player B in round i, player C meets player D in round i,

and player A meets player C in round j, then player B meets player D inround j.

Determine all pairs (n,k) for which there exists an (n,k)-tournament.

12. C6 (COL) A holey triangle is an upward equilateral triangle of side lengthn with n upward unit triangular holes cut out. A diamond is a 60*–120* unitrhombus. Prove that a holey triangle T can be tiled with diamonds if and only ifthe following condition holds: every upward equilateral triangle of side length kin T contains at most k holes, for 1' k' n.

13. C7 (JPN) Consider a convex polyhedron without parallel edges and withoutan edge parallel to any face other than the two faces adjacent to it. Call a pairof points of the polyhedron antipodal if there exist two parallel planes passingthrough these points and such that the polyhedron is contained between theseplanes.Let A be the number of antipodal pairs of vertices, and let B be the number ofantipodal pairs of midpoint edges. Determine the difference A"B in terms ofthe numbers of vertices, edges, and faces.

14. G1 (KOR)IMO1 Let ABC be a triangle with incenter I. A point P in the interiorof the triangle satisfies !PBA+!PCA = !PBC+ !PCB. Show that AP ! AIand that equality holds if and only if P coincides with I.

3.47 IMO 2006 329

15. G2 (UKR) Let ABC be a trapezoid with parallel sides AB >CD. Points K andL lie on the line segments AB and CD, respectively, so that AK/KB= DL/LC.Suppose that there are points P andQ on the line segmentKL satisfying!APB=!BCD and !CQD= !ABC. Prove that the points P, Q, B, andC are concyclic.

16. G3 (USA) Let ABCDE be a convex pentagon such that !BAC = !CAD =!DAE and !ABC = !ACD = !ADE . The diagonals BD and CE meet at P.Prove that the line AP bisects the side CD.

17. G4 (RUS) A point D is chosen on the side AC of a triangle ABC with !C <!A< 90* in such a way that BD= BA. The incircle of ABC is tangent to AB andAC at points K and L, respectively. Let J be the incenter of triangle BCD. Provethat the line KL intersects the line segment AJ at its midpoint.

18. G5 (HEL) In triangle ABC, let J be the center of the excircle tangent to sideBC at A1 and to the extensions of sides AC and AB at B1 and C1, respectively.Suppose that the lines A1B1 and AB are perpendicular and intersect at D. Let Ebe the foot of the perpendicular fromC1 to lineDJ. Determine the angles!BEA1and !AEB1.

19. G6 (BRA) Circles #1 and #2 with centers O1 and O2 are externally tangentat point D and internally tangent to a circle # at points E and F , respectively.Line t is the common tangent of #1 and #2 at D. Let AB be the diameter of #perpendicular to t, so that A, E , and O1 are on the same side of t. Prove that thelines AO1, BO2, EF , and t are concurrent.

20. G7 (SVK) In a triangle ABC, let Ma, Mb, Mc, be respectively the midpoints ofthe sides BC,CA, AB, and let Ta, Tb, Tc be the midpoints of the arcs BC, CA, ABof the circumcircle of ABC, not counting the opposite vertices. For i ( {a,b,c}let #i be the circle withMiTi as diameter. Let pi be the common external tangentto # j, #k ({i, j,k} = {a,b,c}) such that #i lies on the opposite side of pi from# j, #k. Prove that the lines pa, pb, pc form a triangle similar to ABC and find theratio of similitude.

21. G8 (POL) Let ABCD be a convex quadrilateral. A circle passing through thepoints A and D and a circle passing through the points B and C are externallytangent at a point P inside the quadrilateral. Suppose that !PAB+!PDC' 90*and !PBA+!PCD' 90*. Prove that AB+CD! BC+AD.

22. G9 (RUS) Points A1, B1, C1 are chosen on the sides BC, CA, AB of a triangleABC respectively. The circumcircles of triangles AB1C1, BC1A1,CA1B1 intersectthe circumcircle of triangle ABC again at points A2, B2,C2 respectively (A2 &= A,B2 &= B, C2 &= C). Points A3, B3, C3 are symmetric to A1, B1, C1 with respectto the midpoints of the sides BC, CA, AB, respectively. Prove that the trianglesA2B2C2 and A3B3C3 are similar.

23. G10 (SRB)IMO6 Assign to each side b of a convex polygon P the maximumarea of a triangle that has b as a side and is contained in P . Show that the sumof the areas assigned to the sides ofP is at least twice the area ofP .

330 3 Problems

24. N1 (USA)IMO4 Determine all pairs (x,y) of integers satisfying the equation 1+2x+22x+1 = y2.

25. N2 (CAN) For x ( (0,1) let y ( (0,1) be the number whose nth digit after thedecimal point is the 2nth digit after the decimal point of x. Show that if x isrational then so is y.

26. N3 (SAF) The sequence f (1), f (2), f (3), . . . is defined by

f (n) =1n

"#n1

$+

#n2

$+ · · ·+

#nn

$%,

where [x] denotes the integral part of x.(a) Prove that f (n+1) > f (n) infinitely often.(b) Prove that f (n+1) < f (n) infinitely often.

27. N4 (ROU)IMO5 Let P(x) be a polynomial of degree n > 1 with integer co-efficients and let k be a positive integer. Consider the polynomial Q(x) =P(P(. . .P(P(x)) . . . )), where P occurs k times. Prove that there are at most nintegers t such that Q(t) = t.

28. N5 (RUS) Find all integer solutions of the equation

x7"1x"1

= y5"1.

29. N6 (USA) Let a> b> 1 be relatively prime positive integers. Define the weightof an integer c, denoted by w(c), to be the minimal possible value of |x|+ |y|taken over all pairs of integers x and y such that ax+ by = c. An integer c iscalled a local champion if w(c) ! w(c± a) and w(c) ! w(c± b). Find all localchampions and determine their number.

30. N7 (EST) Prove that for every positive integer n there exists an integer m suchthat 2m+m is divisible by n.

730 4 Solutions

4.46 Solutions to the Shortlisted Problems of IMO 2005

1. Clearly, p(x) has to be of the form p(x) = x2+ax±1, where a is an integer. Fora = ±1 and a = 0, polynomial p has the required property: it suffices to takeq= 1 and q= x+1, respectively.Suppose now that |a| ! 2. Then p(x) has two real roots, say x1,x2, which arealso roots of p(x)q(x) = xn+an"1xn"1+ · · ·+a0, ai = ±1. Thus

1=

!!!!an"1xi

+ · · ·+a0xni

!!!! #1|xi|

+ · · ·+1

|xi|n<

1|xi|"1

,

which implies |x1|, |x2| < 2. This immediately rules out the case |a|! 3 and thepolynomials p(x) = x2± 2x" 1. The remaining two polynomials x2± 2x+ 1satisfy the condition for q(x) = x$1.Therefore, the polynomials p(x) with the desired property are x2± x±1, x2±1,and x2±2x+1.

2. Given y > 0, consider the function !(x) = x+ y f (x), x > 0. This function isinjective: indeed, if !(x1) = !(x2), then f (x1) f (y) = f (!(x1)) = f (!(x2)) =f (x2) f (y), so f (x1) = f (x2), so x1 = x2 by the definition of ! . Now if x1 > x2and f (x1) < f (x2), we have !(x1) = !(x2) for y = x1"x2

f (x2)" f (x1)> 0, which

is impossible; hence f is nondecreasing. The functional equation now yieldsf (x) f (y) = 2 f (x+ y f (x)) ! 2 f (x) and consequently f (y) ! 2 for y> 0. There-fore

f (x+ y f (x)) = f (xy) = f (y+ x f (y)) ! f (2x)

holds for arbitrarily small y> 0, implying that f is constant on the interval (x,2x]for each x> 0. But then f is constant on the union of all intervals (x,2x] over allx> 0, that is, on all of R+. Now the functional equation gives us f (x) = 2 for allx, which is clearly a solution.Second Solution. In the same way as abovewe prove that f is nondecreasing, andhence its discontinuity set is at most countable. We can extend f to R%{0} bydefining f (0) = infx f (x) = limx&0 f (x), and the new function f is continuous at0 as well. If x is a point of continuity of f we have f (x) f (0) = limy&0 f (x) f (y) =limy&0 2 f (x+ y f (x)) = 2 f (x), hence f (0) = 2. Now, if f is continuous at 2y,then 2 f (y) = limx&0 f (x) f (y) = limx&0 2 f (x+ y f (x)) = 2 f (2y). Thus f (y) =f (2y), for all but countably many values of y. Being nondecreasing f is a con-stant; hence f (x) = 2.

3. Assume without loss of generality that p! q! r ! s. We have

(pq+ rs)+ (pr+qs)+ (ps+qr)=(p+q+ r+ s)2" p2"q2" r2" s2

2= 30.

It is easy to see that pq+ rs! pr+ qs! ps+ qr, which gives us pq+ rs! 10.Now setting p+ q = x, we obtain x2 + (9" x)2 = (p+ q)2 + (r+ s)2 = 21+2(pq+rs)! 41, which is equivalent to (x"4)(x"5)! 0. Since x= p+q! r+s,we conclude that x! 5. Thus

4.46 Shortlisted Problems 2005 731

25' p2+q2+2pq= 21" (r2+ s2)+2pq' 21+2(pq" rs),

or pq" rs! 2, as desired.Remark. The quadruple (p,q,r,s) = (3,2,2,2) shows that the estimate 2 is thebest possible.

4. Setting y = 0 yields ( f (0) + 1)( f (x)" 1) = 0, and since f (x) = 1 for all x isimpossible, we get f (0) = "1. Now plugging in x = 1 and y = "1 gives usf (1) = 1 or f ("1) = 0. In the first case setting x= 1 in the functional equationyields f (y+1) = 2y+1, i.e., f (x) = 2x"1, which is one solution.Suppose now that f (1) = a &= 1 and f ("1) = 0. Plugging (x,y) = (z,1) and(x,y) = ("z,"1) in the functional equation yields

f (z+1) = (1"a) f (z)+2z+1f ("z"1) = f (z)+2z+1.

It follows that f (z+1)= (1"a) f ("z"1)+a(2z+1), i.e. f (x)= (1"a) f ("x)+a(2x"1). Analogously, f ("x) = (1"a) f (x)+a("2x"1), which together withthe previous equation yields

(a2"2a) f (x) = "2a2x" (a2"2a).

Now a = 2 is clearly impossible. For a &( {0,2} we get f (x) = "2axa"2 " 1. This

function satisfies the requirements only for a = "2, giving the solution f (x) ="x"1. In the remaining case, when a= 0, we have f (x) = f ("x). Setting y= zand y= "z in the functional equation and subtracting yields f (2z) = 4z2"1, sof (x) = x2"1, which satisfies the equation.Thus the solutions are f (x) = 2x"1, f (x) = "x"1, and f (x) = x2"1.

5. The desired inequality is equivalent to

x2+ y2+ z2

x5+ y2+ z2+x2+ y2+ z2

y5+ z2+ x2+x2+ y2+ z2

z5+ x2+ y2' 3. (1)

By the Cauchy inequality we have (x5+ y2+ z2)(yz+ y2+ z2) ! (x5/2(yz)1/2+y2+ z2)2 ! (x2+ y2+ z2)2 and therefore

x2+ y2+ z2

x5+ y2+ z2'yz+ y2+ z2

x2+ y2+ z2.

We get analogous inequalities for the other two summands in (1). Summingthese yields

x2+ y2+ z2

x5+ y2+ z2+x2+ y2+ z2

y5+ z2+ x2+x2+ y2+ z2

z5+ x2+ y2' 2+

xy+ yz+ zxx2+ y2+ z2

,

which together with the well-known inequality x2+ y2+ z2 ! xy+ yz+ zx givesus the result.

732 4 Solutions

Second solution.Multiplying both sides by the common denominator and usingnotation in Chapter 2 (Muirhead’s inequality), we get

T5,5,5+4T7,5,0+T5,2,2+T9,0,0 ! T5,5,2+T6,0,0+2T5,4,0+2T4,2,0+T2,2,2.

By Schur’s and Muirhead’s inequalities we have that T9,0,0+T5,2,2 ! 2T7,2,0 !2T7,1,1. Since xyz! 1 we have that T7,1,1 ! T6,0,0. Therefore

T9,0,0+T5,2,2 ! 2T6,0,0 ! T6,0,0+T4,2,0. (2)

Moreover, Muirhead’s inequality combined with xyz! 1 gives us T7,5,0 ! T5,5,2,2T7,5,0! 2T6,5,1! 2T5,4,0, T7,5,0! T6,4,2! T4,2,0, and T5,5,5! T2,2,2. Adding thesefour inequalities to (2) yields the desired result.

6. A room will be called economic if some of its lamps are on and some are off.Two lamps sharing a switch will be called twins. The twin of a lamp l will bedenoted by l.Suppose we have arrived at a state with the minimum possible number of un-economic rooms, and that this number is strictly positive. Let us choose anyuneconomic room, say R0, and a lamp l0 in it. Let l0 be in a room R1. Switch-ing l0, we make R0 economic; therefore, since the number of uneconomic roomscannot be decreased, this change must make room R1 uneconomic. Now choosea lamp l1 in R1 having the twin l1 in a room R2. Switching l1 makes R1 economic,and thus must make R2 uneconomic. Continuing in this manner we obtain a se-quence l0, l1, . . . of lamps with li in a room Ri and li &= li+1 in Ri+1 for all i. Thelamps l0, l1, . . . are switched in this order. This sequence has the property thatswitching li and li makes room Ri economic and room Ri+1 uneconomic.Let Rm = Rk with m > k be the first repetition in the sequence (Ri). Let us stopswitching the lamps at lm"1. The room Rk was uneconomic prior to switchinglk. Thereafter, lamps lk and lm"1 have been switched in Rk, but since these twolamps are distinct (indeed, their twins lk and lm"1 are distinct), the room Rk isnow economic, as well as all the rooms R0, R1, . . . , Rm"1. This decreases thenumber of uneconomic rooms, contradicting our assumption.

7. Let v be the number of video winners. One easily finds that for v= 1 and v= 2,the number n of customers is at least 2k+ 3 and 3k+ 5 respectively. We proveby induction on v that if n! k+1, then n! (k+2)(v+1)"1.Without loss of generality, we can assume that the total number n of customersis minimum possible for given v > 0. Consider a person P who was convincedby nobody but himself. Then P must have won a video; otherwise, P could beremoved from the group without decreasing the number of video winners. LetQ and R be the two persons convinced by P. We denote by C the set of personsinfluenced by P through Q to buy a sombrero, including Q, and by D the setof all other customers excluding P. Let x be the number of video winners in C .Then there are v" x"1 video winners in D . We have |C |! (k+2)(x+1)"1,by the induction hypothesis if x> 0 and because P is a winner if x= 0. Similarly,|D | ! (k+ 2)(v" x)" 1. Thus n ! 1+(k+ 2)(x+ 1)" 1+(k+ 2)(v" x)" 1,i.e., n! (k+2)(v+1)"1.


8. Suppose that a two-sided m$ n board T is considered, where exactly k of thesquares are transparent. A transparent square is colored only on one side (thenit looks the same from the other side), while a nontransparent one needs to becolored on both sides, not necessarily in the same color.LetC =C(T ) be the set of colorings of the board in which there exist two blackpaths from the left edge to the right edge, one on top and one underneath, notintersecting at any transparent square. If k = 0 then |C| = N2. We prove by in-duction on k that 2k|C|'N2. This will imply the statement of the problem, since|C| =M for k = mn.Let q be a fixed transparent square. Consider any coloring B in C: If q is con-verted into a nontransparent square, a new board T + with k" 1 transparentsquares is obtained, so by the induction hypothesis 2k"1|C(T +)| ' N2. SinceB contains two black paths at most one of which passes through q, color-ing q in either color on the other side will result in a coloring in C+; hence|C(T +)|! 2|C(T )|, implying 2k|C(T )|' N2 and finishing the induction.Second solution. By a path we shall mean a black path from the left edge tothe right edge. Let A denote the set of pairs of m$ n boards each of whichhas a path. Let B denote the set of pairs of boards such that the first boardhas two nonintersecting paths. Obviously, |A | = N2 and |B| = 2mnM. To prove|A |! |B|, we will construct an injection f :B # A .Among paths on a given board we define path x to be lower than y if the set ofsquares “under” x is a subset of the squares under y. This relation is a relationof incomplete order. However, for each board with at least one path there existsa lowest path (comparing two intersecting paths, we can always take the “lowerbranch” on each nonintersecting segment). Now, for a given element of B, we“swap” the lowest path and all squares underneath on the first board with thecorresponding points on the other board. This swapping operation is the desiredinjection f . Indeed, since the first board still contains the highest path (whichdidn’t intersect the lowest one), the new configuration belongs to A . On theother hand, this configuration uniquely determines the lowest path on the originalelement ofB; hence no two different elements ofB can go to the same elementof A . This completes the proof.

9. Let [XY ] denote the label of segment XY , where X and Y are vertices of thepolygon. Consider any segment MN with the maximum label [MN] = r. Bycondition (ii), for any Pi &= M,N, exactly one of PiM and PiN is labeled by r.Thus the set of all vertices of the n-gon splits into two complementary groups:A = {Pi | [PiM] = r} and B = {Pi | [PiN] = r}. We claim that a segment XYis labelled by r if and only if it joins two points from different groups. Assumewithout loss of generality that X ( A . If Y ( A , then [XM] = [YM] = r, so[XY ] < r. If Y ( B, then [XM] = r and [YM] < r, so [XY ] = r by (ii), as weclaimed.We conclude that a labeling satisfying (ii) is uniquely determined by groups A

andB and labelings satisfying (ii) within A and B.

734 4 Solutions

(a) We prove by induction on n that the greatest possible value of r is n" 1.The degenerate cases n = 1,2 are trivial. If n ! 3, the number of differentlabels of segments joining vertices inA (resp.B) does not exceed |A |"1(resp. |B|"1), while all segments joining a vertex in A and a vertex inB

are labeled by r. Therefore r ' (|A |"1)+(|B|"1)+1= n"1. Equalityis achieved if all the mentioned labels are different.

(b) Let an be the number of labelings with r = n" 1. We prove by inductionthat an = n!(n"1)!

2n"1 . This is trivial for n= 1, so let n! 2. If |A | = k is fixed,the groups A and B can be chosen in

&nk'ways. The set of labels used

within A can be selected among 1,2, . . . ,n" 2 in&n"2k"1

'ways. Now the

segments within groups A and B can be labeled so as to satisfy (ii) in akand an"k ways, respectively. In this way, every labeling has been countedtwice, since choosingA is equivalent to choosingB. It follows that

an =12

n"1

$k=1

(nk

)(n"2k"1

)akan"k

=n!(n"1)!2(n"1)

n"1

$k=1

akk!(k"1)! ·

an"k(n" k)!(n" k"1)!

=n!(n"1)!2(n"1)

n"1

$k=1

12k"1

·1

2n"k"1=n!(n"1)!2n"1

.

10. Denote by L the leftmost and by R the rightmost marker. To start with, note thatthe parity of the number of black-side-up markers remains unchanged. Hence, ifonly two markers remain, these markers must have the same color up.We shall show by induction on n that the game can be successfully finished ifand only if n % 0 or n % 2 (mod 3), and that the upper sides of L and R will beblack in the first case and white in the second case.The statement is clear for n = 2,3. Assume that we have finished the game forsome n, and denote by k the position of the marker X (counting from the left) thatwas last removed. Having finished the game, we have also finished the subgameswith the k markers from L to X and with the n" k+ 1 markers from X to R(inclusive). Thereby, before X was removed, the upper side of L had been blackif k % 0 and white if k % 2 (mod 3), while the upper side of R had been black ifn"k+1% 0 and white if n"k+1% 2 (mod 3). Markers L and R were reversedupon the removal of X . Therefore, in the final position, L and R are white if andonly if k % n" k+ 1 % 0, which yields n % 2 (mod 3), and black if and only ifk % n" k+1% 2, which yields n% 0 (mod 3).On the other hand, a game with n markers can be reduced to a game with n"3markers by removing the second, fourth, and third markers in this order. Thisfinishes the induction.Second solution. An invariant can be defined as follows. To each white markerwith k black markers to its left we assign the number ("1)k. Let S be the sum ofthe assigned numbers. Then it is easy to verify that the remainder of S modulo


3 remains unchanged throughout the game: For example, when a white markerwith two white neighbors and k black markers to its left is removed, S decreasesby 3("1)t .Initially, S = n. In the final position with two markers remaining, S equals 0 ifthe two markers are black and 2 if these are white (note that, as before, the twomarkers must be of the same color). Thus n% 0 or 2 (mod 3).Conversely, a game with n markers is reduced to n" 3 markers as in the firstsolution.

11. Assume that there were n contestants, ai of whom solved exactly i problems,where a0+ · · ·+a5 = n. Let us count the number N of pairs (C,P), where con-testant C solved the pair of problems P. Each of the 15 pairs of problems wassolved by at least 2n+15 contestants, implying N ! 15 · 2n+15 = 6n+ 3. On theother hand, ai students solved i(i"1)

2 pairs; hence

6n+3' N ' a2+3a3+6a4+10a5 = 6n+4a5" (3a3+5a2+6a1+6a0).

Consequently a5 ! 1. Assume that a5 = 1. Then we must have N = 6n+ 4,which is possible only if 14 of the pairs of problems were solved by exactly2n+15 students and the remaining one by 2n+1

5 + 1 students, and all students butthe winner solved 4 problems.The problem t not solved by the winner will be called tough and the pair ofproblems solved by 2n+1

5 +1 students special.Let us count the numberMp of pairs (C,P) for which P contains a fixed problemp. Let bp be the number of contestants who solved p. Then Mt = 3bt (each ofthe bt students solved three pairs of problems containing t), and Mp = 3bp+ 1for p &= t (the winner solved four such pairs). On the other hand, each of the fivepairs containing p was solved by 2n+1

5 or 2n+15 + 1 students, so Mp = 2n+ 2 ifthe special pair contains p, andMp = 2n+1 otherwise.Now since Mt = 3bt = 2n+ 1 or 2n+ 2, we have 2n+ 1% 0 or 2 (mod 3). Butif p &= t is a problem not contained in the special pair, we haveMp = 3bp+1=2n+1; hence 2n+1% 1 (mod 3), which is a contradiction.

12. Suppose that there exist desired permutations ! and " for some sequencea1, . . . ,an. Given a sequence (bi) with sum divisible by n that differs modulo nfrom (ai) in only two positions, say i1 and i2, we show how to construct desiredpermutations ! + and " + for sequence (bi). In this way, starting from an arbitrarysequence (ai) for which ! and " exist, we can construct desired permutations forany other sequence with sum divisible by n. All congruences below are modulon.We know that !(i) + "(i) % bi for all i &= i1, i2. We construct the sequencei1, i2, i3, . . . as follows: for each k ! 2, ik+1 is the unique index such that

!(ik"1)+ "(ik+1) % bik . (1)

Let ip = iq be the repetition in the sequence with the smallest q. We claim thatp= 1 or p= 2. Assume to the contrary that p> 2. Summing (1) for k= p, p+1,

736 4 Solutions

. . . , q"1 and taking the equalities !(ik)+"(ik) = bik for ik &= i1, i2 into account,we obtain !(ip"1)+!(ip)+"(iq"1)+"(iq)% bp+bq"1. Since iq = ip, it followsthat !(ip"1)+ "(iq"1) % bq"1 and therefore ip"1 = iq"1, a contradiction. Thusp= 1 or p = 2 as claimed.Now we define the following permutations:

! +(ik) = !(ik"1) for k = 2,3, . . . ,q"1 and ! +(i1) = !(iq"1),

" +(ik) = "(ik+1) for k= 2,3, . . . ,q"1 and " +(i1) =

*"(i2) if p= 1,"(i1) if p= 2;

! +(i) = !(i) and " +(i) = "(i) for i &( {i1, . . . , iq"1}.

Permutations ! + and " + have the desired property. Indeed, ! +(i)+ " +(i) = bi ob-viously holds for all i &= i1, but then it must also hold for i= i1.

13. For every green diagonal d, let Cd denote the number of green–red intersectionpoints on d. The task is to find the maximum possible value of the sum $dCdover all green diagonals.Let di and d j be two green diagonals and let the part of polygonM lying betweendi and d j have m vertices. There are at most n"m"1 red diagonals intersectingboth di and d j, while each of the remaining m" 2 diagonals meets at most oneof di,d j. It follows that

Cdi +Cdj ' 2(n"m"1)+ (m"2)= 2n"m"4. (1)

We now arrange the green diagonals in a sequence d1,d2, . . . ,dn"3 as follows.It is easily seen that there are two green diagonals d1 and d2 that divide M intotwo triangles and an (n" 2)-gon; then there are two green diagonals d3 and d4that divide the (n"2)-gon into two triangles and an (n"4)-gon, and so on. Wecontinue this procedure until we end up with a triangle or a quadrilateral. Now,the part of M between d2k"1 and d2k has at least n" 2k vertices for 1 ' k 'r, where n" 3 = 2r+ e, e ( {0,1}; hence, by (1), Cd2k"1 +Cd2k ' n+ 2k" 4.Moreover,Cdn"3 ' n"3. Summing yields

Cd1 +Cd2 + · · ·+Cdn"3 'r

$k=1

(n+2k"4)+ e(n"3)

= 3r2+ e(3r+1) =

+34(n"3)2

,.

This value is attained in the following example. Let A1A2 . . .An be the n-gon Mand let l =

- n2.+ 1. The diagonals A1Ai, i = 3, . . . , l, and AlA j, j = l+ 2, . . . ,n

are colored green, whereas the diagonals A2Ai, i= l+1, . . . ,n, and Al+1Aj, j =3, . . . , l"1 are colored red.Thus the answer is , 34 (n"3)

2-.

14. Let F be the point of tangency of the incircle with AC and let M and N be therespective points of tangency of AB and BC with the corresponding excircles. IfI is the incenter and Ia and P respectively the center and the tangency point withray AC of the excircle corresponding to A, we have AI

IL = AIIF = AIa

IaP = AIaIaN , which


implies that.AIL/.AIaN. Thus L lies on AN, and analogouslyK lies onCM.Define x = AF and y = CF . Since BD = BE , AD = BM = x, and CE = BN =y, the condition AB+BC = 3AC gives us DM = y and EN = x. The trianglesCLN andMKA are congruent since their altitudes KD and LE satisfy DK = EL,DM =CE , and AD= EN. Thus!AKM = !CLN, implying that ACKL is cyclic.

15. Let P be the fourth vertex of the rhombusC2A1A2P. Since.C2PC1 is equilateral,we easily conclude that B1B2C1P is also a rhombus. Thus.PB1A2 is equilateraland !(C2A1,C1B2) = !A2PB1 = 60*. It easily follows that.AC1B2 /= .BA1C2and consequently AC1 = BA1; similarly, BA1 =CB1. Therefore triangle A1B1C1is equilateral. Now it follows from B1B2 = B2C1 that A1B2 bisects !C1A1B1.Similarly, B1C2 and C1A2 bisect !A1B1C1 and !B1C1A1; hence A1B2, B1C2,C1A2 meet at the incenter of A1B1C1, i.e. at the center of ABC.

16. Since !ADL= !KBA= 180*" 12!BCD and !ALD= 1

2!AYD= !KAB, trian-gles ABK and LDA are similar. Thus BK

BC = BKAD = AB

DL = DCDL , which together with

!LDC= !CBK gives us.LDC/.CBK. Therefore!KCL= 360*"!BCD"(!LCD+!KCB) = 360*"!BCD" (!CKB+!KCB) = 180*"!CBK, whichis constant.

17. To start with, we note that points B,E,C are the images of D,F,A respect-ively under the rotation around point O for the angle # = !DOB, where O isthe intersection of the perpendicular bisectors of AC and BD. Then OE = OFand !OFE = !OAC = 90" #

2 ; hence the points A,F,R,O are on a circleand !ORP = 180* "!OFA. Analogously, the points B,E,Q,O are on a cir-cle and !OQP = 180* "!OEB = !OEC = !OFA. This shows that !ORP =180*"!OQP, i.e. the point O lies on the circumcircle of.PQR, thus being thedesired point.

18. Let O and O1 be the circumcenters of triangles ABC and ADE , respectively.It is enough to show that HM 0 OO1. Let AA+ be the diameter of the cir-cumcircle of ABC. We note that if B1 is the foot of the altitude from B, thenHE bisects !CHB1. Since the triangles COM and CHB1 are similar (indeed,!CHB= !COM = !A), we have CE

EB1 = CHHB1 = CO

OM = 2COAH = A+A

AH .Thus, if Q is the intersection pointof the bisector of !A+AH with HA+,we obtain CE

EB1 = A+QQH , which together

with A+C 1 AC and HB1 1 AC givesus QE 1 AC. Analogously, QD 1 AB.Therefore AQ is a diameter of the cir-cumcircle of.ADE andO1 is the mid-point of AQ. It follows that OO1 is theline passing through the midpoints ofAQ and AA+; hence OO10HM.

A

B C

D

E

H

M

O

A+

Q

B1

O1

Second solution.We again prove that OO1 0 HM. Since AA+ = 2AO, it sufficesto prove AQ= 2AO1.

738 4 Solutions

Elementary calculations of angles give us!ADE = !AED= 90*" %2 . Applying

the law of sines to.DAH and.EAH we now haveDE =DH+EH= AH cos&cos %2

+AH cos'cos %2

. Since AH = 2OM = 2Rcos% , we obtain

AO1 =DE2sin%

=AH(cos& + cos')2sin% cos %2

=2Rcos% sin %2 cos(

&"'2 )

sin% cos %2.

We now calculate AQ. Let N be the intersection of AQ with the circumcircle.Since!NAO= &"'

2 , we have AN = 2Rcos(&"'2 ). Noting that.QAH /.QNM(and that MN = R"OM), we have

AQ=AN ·AHMN+AH

=2Rcos(&"'2 ) ·2cos%

1+ cos%=2Rcos(&"'2 )cos%

cos2 %2= 2AO1.

19. We denote byD,E,F the points of tan-gency of the incircle with BC,CA,AB,respectively, by I the incenter, and byY + the intersection of AX and LY . SinceEF is the polar line to the point Awith respect to the incircle, it meetsAL at point R such that A,R;K,L areconjugate, i.e., KRRL = KA

AL . ThenKXLY + =

KAAL = KR

RL = KXLY and therefore LY =

LY , where Y is the intersection of XRand LY . Thus showing that LY = LY +

A

B CD

E

FI

R

K

L

MY

X

Y +

I+

PQ

(which is the same as showing that PM = MQ, i.e., CP = QB) is equivalent toshowing that XY contains R. Since XKYL is an inscribed trapezoid, it is enoughto show that R lies on its axis of symmetry, that is, DI.Since AM is the median, the triangles ARB and ARC have equal areas, and since!(RF,AB) = !(RE,AC) we have that 1 =

S.ABRS.ACR

= (AB·FR)(AC·ER) . Hence

ABAC = ER

FR .Let I+ be the point of intersection of the line through F parallel to IE with theline IR. Then FI+

EI = FRRE = AC

AB and !I+FI = !BAC (angles with orthogonal rays).Thus the triangles ABC and FII+ are similar, implying that!FII+ = !ABC. Since!FID= 180*"!ABC, it follows that R, I, and D are collinear.

20. We shall prove the inequalities p(ABC) ! 2p(DEF) and p(PQR) ! 12 p(DEF).

The statement of the problem will immediately follow.Let Db and Dc be the reflections of D in AB and AC, and let A1,B1,C1 bethe midpoints of BC,CA,AB, respectively. It is easy to see that Db,F,E,Dc arecollinear. Hence p(DEF) =DbF+FE+EDc =DbDc'DbC1+C1B1+B1Dc =12 (AB+BC+CA) = 1

2 p(ABC).To prove the second inequality we observe that P, Q, and R are the points oftangency of the excircles with the sides of .DEF . Let FQ = ER = x, DR =FP = y, and DQ = EP = z, and let ( ,),* be the angles of .DEF at D,E,F ,


respectively. Let Q+ and R+ be the projections of Q and R onto EF , respectively.Then QR ! Q+R+ = EF " FQ+ " R+E = EF " x(cos* + cos)). Summing thiswith the analogous inequalities for FD and DE , we obtain

p(PQR) ! p(DEF)" x(cos*+ cos))" y(cos( + cos*)" z(cos( + cos)).

Assuming without loss of generality that x' y' z, we also haveDE ' FD'FEand consequently cos* + cos) ! cos( + cos* ! cos( + cos) . Now Cheby-shev’s inequality gives us p(PQR) ! p(DEF)" 2

3 (x+ y+ z)(cos) + cos* +

cos( )! p(DEF)"(x+y+z)= 12 p(DEF), wherewe used x+y+z= 1

2 p(DEF)and the fact that the sum of the cosines of the angles in a triangle does not exceed32 . This finishes the proof.

21. We will show that 1 is the only such number. It is sufficient to prove that forevery prime number p there exists some am such that p | am. For p = 2,3 wehave p | a2 = 48. Assume now that p> 3. Applying Fermat’s theorem, we have

6ap"2 = 3 ·2p"1+2 ·3p"1+6p"1"6% 3+2+1"6= 0 (mod p).

Hence p | ap"2, i.e. gcd(p,ap"2) = p> 1. This completes the proof.

22. It immediately follows from the condition of the problem that all the terms ofthe sequence are distinct. We also note that |ai"an|' n"1 for all integers i,nwhere i< n, because if d = |ai"an|! n then {a1, . . . ,ad} contains two elementscongruent to each other modulo d, which is a contradiction. It easily followsby induction that for every n ( N the set {a1, . . . ,an} consists of consecutiveintegers. Thus, if we assumed that some integer k did not appear in the sequencea1,a2, . . . , the same would have to hold for all integers either larger or smallerthan k, which contradicts the condition that infinitely many positive and negativeintegers appear in the sequence. Thus, the sequence contains all integers.

23. Let us consider the polynomial

P(x) = (x+a)(x+b)(x+ c)" (x"d)(x" e)(x" f )= Sx2+Qx+R,

where Q= ab+bc+ ca"de" e f" f d and R= abc+de f .Since S | Q,R, it follows that S | P(x) for every x ( Z. Hence, S | P(d) = (d+a)(d + b)(d+ c). Since S > d+ a, d + b, d + c and thus cannot divide any ofthem, it follows that S must be composite.

24. We will show that n has the desired property if and only if it is prime.For n = 2 we can take only a = 1. For n > 2 and even, 4 | n!, but an + 1 %1,2 (mod 4), which is impossible. Now we assume that n is odd. Obviously(n!"1)n+1% ("1)n+1= 0 (mod n!). If n is composite and d its prime divisor,then

& n!d "1

'n+1=$nk=1

&nk' n!kdk , where each summand is divisible by n! because

d2 | n!; therefore n! divides& n!d "1

'n+1. Thus, all composite numbers are ruled

out.It remains to show that if n is an odd prime and n! | an+ 1, then n! | a+ 1, andtherefore a= n!"1 is the only relevant value for which n! | an+1. Consider any

740 4 Solutions

prime number p' n. If p | an+1a+1 , we have p | ("a)n"1 and by Fermat’s theorem

p | ("a)p"1" 1. Therefore p | ("a)(n,p"1)" 1 = "a" 1, i.e. a % "1 (mod p).But then an+1

a+1 = an"1"an"2+ · · ·"a+1% n (mod p), implying that p = n. Itfollows that an+1a+1 is coprime to (n"1)! and consequently (n"1)! divides a+1.Moreover, the above consideration shows that nmust divide a+1. Thus n! | a+1as claimed. This finishes our proof.

25. We will use the abbreviation HD to denote a “highly divisible integer.” Letn = 2%2(n)3%3(n) · · · p%p(n) be the factorization of n into primes. We have d(n) =(%2(n)+1) · · ·(%p(n)+1). We start with the following two lemmas.Lemma 1. If n is an HD and p,q primes with pk < ql (k, l ( N), then

k%q(n) ' l%p(n)+ (k+1)(l"1).

Proof. The inequality is trivial if %q(n) < l. Suppose that %q(n) ! l. Thennpk/ql is an integer less than q, and d(npk/ql) < d(n), which is equivalentto (%q(n)+ 1)(%p(n)+ 1) > (%q(n)" l+ 1)(%p(n)+ k+ 1) implying thedesired inequality.

Lemma 2. For each p and k there exist only finitely many HD’s n such that%p(n) ' k.

Proof. It follows from Lemma 1 that if n is an HD with %p(n) ' k, then %q(n)is bounded for each prime q and %q(n) = 0 for q > pk+1. Therefore thereare only finitely many possibilities for n.

We are now ready to prove both parts of the problem.(a) Suppose that there are infinitely many pairs (a,b) of consecutive HD’s with

a | b. Since d(2a) > d(a), we must have b= 2a. In particular, d(s) ' d(a)for all s < 2a. All but finitely many HD’s a are divisible by 2 and by 37.Then d(8a/9) < d(a) and d(3a/2) < d(a) yield

(%2(a)+4)(%3(a)"1) < (%2(a)+1)(%3(a)+1)2 3%3(a)"5< 2%2(a),%2(a)(%3(a)+2)' (%2(a)+1)(%3(a)+1)2 %2(a) ' %3(a)+1.

We now have 3%3(a)"5< 2%2(a) ' 2%3(a)+22 %3(a) < 7, which is acontradiction.

(b) Assume for a given prime p and positive integer k that n is the smallest HDwith %p ! k. We show that np is also an HD. Assume the opposite, i.e., thatthere exists an HD m< n

p such that d(m) ! d( np). By assumption, m mustalso satisfy %p(m)+1' %p(n). Then

d(mp) = d(m)%p(m)+2%p(m)+1

! d(np

)%p(n)+1%p(n)

= d(n),

contradicting the initial assumption that n is an HD (since mp < n). Thisproves that np is an HD. Since this is true for every positive integer k, theproof is complete.


26. Assuming b &= a, it trivially follows that b> a. Let p> b be a prime number andlet n= (a+1)(p"1)+1.We note that n% 1 (mod p"1) and n%"a (mod p). Itfollows that rn = r ·(rp"1)a+1% r (mod p) for every integer r. We now have an+n% a"a= 0 (mod p). Thus, an+n is divisible by p, and hence by the conditionof the problem bn + n is also divisible by p. However, we also have bn + n %b"a (mod p), i.e., p | b"a, which contradicts p> b. Hence, it must follow thatb = a. We note that b = a trivially fulfills the conditions of the problem for alla ( N.

27. Let p be a prime and k < p an even number. We note that (p" k)!(k" 1)! %("1)k"1(p" k)!(p" k+1) · · ·(p"1) = ("1)k"1(p"1)!% 1 (mod p) by Wil-son’s theorem. Therefore

(k"1)!nP((p" k)!) = $ni=0 ai[(k"1)!]n"i[(p" k)!(k"1)!]i% $ni=0 ai[(k"1)!]n"i = S((k"1)!) (mod p),

where S(x) = an + an"1x+ · · ·+ a0xn. Hence p | P((p" k)!) if and only if p |S((k" 1)!). Note that S((k" 1)!) depends only on k. Let k > 2an + 1. Then,s= (k"1)!/an is an integer that is divisible by all primes smaller than k. HenceS((k"1)!) = anbk for some bk % 1 (mod s). It follows that bk is divisible only byprimes larger than k. For large enough k we have |bk| > 1. Thus for every primedivisor p of bk we have p | P((p" k)!).It remains to select a large enough k for which |P((p" k)!)| > p. We take k =(q"1)!, where q is a large prime. All the numbers k+ i for i= 1,2, . . . ,q"1 arecomposite (by Wilson’s theorem, q | k+1). Thus p = k+q+ r, for some r ! 0.We now have |P((p" k)!)|= |P((q+ r)!)|> (q+ r)!> (q"1)!+q+ r= p, forlarge enough q, since n= degP! 2. This completes the proof.Remark. The above solution actually also works for all linear polynomials Pother than P(x) = x+a0. Nevertheless, these particular cases are easily handled.If |a0| > 1, then P(m!) is composite for m > |a0|, whereas P(x) = x+ 1 andP(x) = x" 1 are both composite for, say, x = 5!. Thus the condition n ! 2 wasredundant.

742 4 Solutions

4.47 Solutions to the Shortlisted Problems of IMO 2006

1. If a0! 0 then ai! 0 for each i and [ai+1]' ai+1 = [ai]{ai}< [ai] unless [ai] = 0.Eventually 0 appears in the sequence [ai] and all subsequent ak’s are 0.Now suppose that a0 < 0; then all ai ' 0. Suppose that the sequence neverreaches 0. Then [ai] ' "1 and so 1+ [ai+1] > ai+1 = [ai]{ai} > [ai], so the se-quence [ai] is nondecreasing and hence must be constant from some term on:[ai] = c < 0 for i ! n. The defining formula becomes ai+1 = c{ai} = c(ai" c),which is equivalent to bi+1 = cbi, where bi = ai" c2

c"1 . Since (bi) is bounded,we must have either c= "1, in which case ai+1 = "ai"1 and hence ai+2 = ai,or bi = 0 and thus ai = c2

c"1 for all i! n.

2. We use induction on n. We have a1 = 1/2; assume that n! 1 and a1, . . . ,an > 0.The formula gives us (n+ 1)$mk=1

akm"k+1 = 1. Writing this equation for n and

n+1 and subtracting yields

(n+2)an+1 =n

$k=1

(n+1

n" k+1"

n+2n" k+2

)ak,

which is positive, as is the coefficient at each ak.Remark. Using techniques from complex analysis such as contour integrals, onecan obtain the following formula for n! 1:

an =/ +

1

dxxn(,2+ ln2(x"1))

> 0.

3. We know that cn = -n"1".n"1-". , where - = 1+

)5

2 and . = 1")5

2 are the roots oft2" t"1. Since cn"1/cn#". , taking % =. and & = 1 is a natural choice. Forevery finite set J 3 N we have

"1=+

$n=0

.2n+1 < .x+ y= $j(J. j"1 <

+

$n=0

.2n = - .

Thusm="1 andM= - is an appropriate choice. We now prove that this choicehas the desired properties by showing that for any x,y ( N with "1< K = x.+y< - , there is a finite set J 4 N such that K = $ j(J.

j.Given such K, there are sequences i1 ' · · · ' ik with . i1 + · · ·+. ik = K (onesuch sequence consists of y zeros and x ones). Consider all such sequences ofminimum length n. Since .m+.m+1 = .m+2, these sequences contain no twoconsecutive integers. Order such sequences as follows: If ik = jk for 1' k' t andit < jt , then (ir) 5 ( jr). Consider the smallest sequence (ir)nr=1 in this ordering.We claim that its terms are distinct. Since 2.2 = 1+.3, replacing two equaltermsm,m bym"2,m+1 form! 2 would yield a smaller sequence, so only 0 or1 can repeat among the ir. But it = it+1 = 0 implies$r.

ir > 2+$+k=0.2k+3 = - ,

while it = it+1 = 1 similarly implies $r. ir < "1, so both cases are impossible,proving our claim. Thus J = {i1, . . . , in} is a required set.


4. Since aba+b = 1

4

"a+b" (a"b)2

a+b

%, the left hand side of the desired inequality

equals

A=$i< j

aia jai+a j

=n"14 $

kak"

14$i< j

(ai"a j)2

ai+a j.

The righthand side of the inequality is equal to

B=n2$aia j$ak

=n"14 $

kak"

14$i< j

(ai"a j)2

$ak.

Now A' B follows from the trivial inequality $ (ai"a j)2ai+a j ! $

(ai"a j)2

$ak.

5. Let x=)b+

)c"

)a, y=

)c+

)a"

)b, and z=

)a+

)b"

)c. All of these

numbers are positive because a,b,c are sides of a triangle. Then b+ c" a =x2" 1

2(x" y)(x" z) and)b+ c"a)

b+)c"

)a

=

01"

(x" y)(y" z)2x2

' 1"(x" y)(x" z)

4x2.

Now it is enough to prove that

x"2(x" y)(x" z)+ y"2(y" z)(y" x)+ z"2(z" x)(z" y)! 0,

which directly follows from Schur’s inequality.

6. Assume, without loss of generality, that a ! b ! c. The lefthand side of theinequality equals L = (a" b)(b" c)(a" c)(a+ b+ c). From (a" b)(b" c) '14 (a"c)2 we get L' 1

4(a"c)3|a+b+c|. The inequality (a"c)2 ' 2(a"b)2+2(b" c) implies (a" c)2 ' 2

3 [(a"b)2+(b" c)2+(a" c)2]. Therefore

L')22

((a"b)2+(b" c)2+(a" c)2

3

)3/2(a+b+ c).

Finally, the mean inequality gives us

L ')22

((a"b)2+(b" c)2+(a" c)2+(a+b+ c)2

4

)2

=9)2

32(a2+b2+ c2)2.

Equality is attained if and only if a" b = b" c and (a" b)2+(b" c)2+(a"c)2 = 3(a+ b+ c)2, which leads to a =

"1+ 3)

2

%b and c =

"1" 3)

2

%b. Thus

M = 9)2

32 .Second solution. We have L = |(a" b)(b" c)(c" a)(a+ b+ c)|. Without lossof generality, assume that a+ b+ c = 1 (the case a+ b+ c = 0 is trivial). Themonic cubic polynomial with roots a"b, b" c, and c"a is of the form

744 4 Solutions

P(x) = x3+qx+ r, q=12"32(a2+b2+ c2), r = "(a"b)(b" c)(c"a).

Then M2 = maxr2/"1"2q3

%4. Since P(x) has three real roots, its discrimi-

nant (q/3)3 + (r/2)2 must be positive, so r2 ! " 427q

3. Thus M2 ' f (q) =

" 427q

3/"1"2q3

%4. The function f attains its maximum 34/29 at q = "3/2, so

M ' 9)2

32 . The case of equality is easily computed.Third solution. Assume that a2+b2+c2 = 1 and write u= (a+b+c)/

)3, v=

(a+ )b+ )2c)/)3, w= (a+ )2b+ )c)/

)3, where ) = e2, i/3. Then analogous

formulas hold for a,b,c in terms of u,v,w, fromwhich one directly obtains |u|2+|v|2+ |w|2 = a2+b2+ c2 = 1 and

a+b+ c=)3u, |a"b|= |v" )w|, |a" c|= |v" )2w|, |b" c|= |v"w|.

Thus L =)3|u||v3"w3|'

)3|u|(|v|3+ |w|3) '

132 |u|2(1" |u|2)3 ' 9

)2

32 . It iseasy to trace back a,b,c to the equality case.

7. (a) We show that for n = 2k all lamps will be switched on in n" 1 steps andoff in n steps. For k = 1 the statement is true. Suppose it holds for some kand let n= 2k+1; define L= {L1, . . . ,L2k} and R= {L2k+1, . . . ,L2k+1}. Thefirst 2k"1 steps are performed without any influence on or from the lampsfrom R; thus after 2k"1 steps the lamps in L are on and those from R areoff. After the 2kth step, L2k and L2k+1 are on and the other lamps are off.Notice that from now on, L and R will be symmetric (i.e., Li and L2k+1"iwill have the same state) and will never influence each other. Since R startswith only the leftmost lamp on, in 2k steps all its lamps will be off. Thesame will happen to L. There are 2k+2k = 2k+1 steps in total.

(b) We claim that for n = 2k + 1 the lamps cannot be switched off. After thefirst step, only L1 and L2 are on. According to (a), after 2k " 1 steps alllamps but Ln will be on, so after the 2kth step all lamps will be off exceptfor Ln"1 and Ln. Since this position is symmetric to the one after the firststep, the procedure will never end.

8. We call a triangle odd if it has two odd sides. To any odd isosceles triangleAiA jAk we assign a pair of sides of the 2006-gon. We may assume that k" j =j" i > 0 is odd. A side of the 2006-gon is said to belong to triangle AiA jAkif it lies on the polygonal line AiAi+1 . . .Ak. At least one of the odd number ofsides AiAi+1, . . . ,Aj"1Aj and at least one of the sides AjA j+1, . . . ,Ak"1Ak do notbelong to any other odd isosceles triangle; assign those two sides to .AiA jAk.This ensures that every two assigned pairs are disjoint; therefore there are atmost 1003 odd isosceles triangles.An examplewith 1003 odd isosceles triangles can be attained when the diagonalsA2kA2k+2 are drawn for k = 0, . . . ,1002, where A0 = A2006.

9. The number c(P) of points inside P is equal to n" a(P)" b(P), where n = |S|.Writing y= 1" x, the considered sum becomes


$Pxa(P)yb(P)(x+ y)c(P) =$

P

c(P)

$i=0

(c(P)

i

)xa(P)+iyb(P)+c(P)"i

=$P

a(P)+c(P)

$k=a(P)

(c(P)

k"a(P)

)xkyn"k.

Here the coefficient at xkyn"k is the sum $P& c(P)k"a(P)

', which equals the number

of pairs (P,Z) of a convex polygon P and a k-element subset Z of S whose con-vex hull is P, and is thus equal to

&nk'. Now the required statement immediately

follows.

10. Denote by SA (R) the number of strawberries of arrangementA inside rectangleR. We writeA ' B if for every rectangle Q containing the top left corner O wehave SB(Q) ! SA (Q). In this ordering, every switch transforms an arrangementto a larger one. Since the number of arrangements is finite, it is enough to provethat whenever A < B there is a switch taking A to C with C ' B. Considerthe highest row t of the cake that differs inA andB; let X andY be the positionsof the strawberries in t in A and B respectively. Clearly Y is to the left fromX and the strawberry of A in the column of Y is below Y . Now consider thehighest strawberryX + ofA below t whose column is between X andY (includingY ). Let s be the row of X +. Now switch X ,X + to the other two vertices Z,Z+ ofthe corresponding rectangle, obtain-ing an arrangement C . We claimthat C ' B. It is enough to ver-ify that SC (Q) ' SB(Q) for thoserectangles Q = OMNP with N ly-ing inside XZX +Z+. Let Q+ = OMN1P1be the smallest rectangle contain-ing X . Our choice of s ensures thatSC (Q) = SA (Q+) ! SB(Q+) ! SB(Q),as claimed.

X +

O

M NN1

P P1

t

s

XY Z

Z+

11. Let q be the largest integer such that 2q | n. We prove that an (n,k)-tournamentexists if and only if k< 2q.The first l rounds of an (n,k)-tournament form an (n, l)-tournament. Thus it isenough to show that an (n,2q"1)-tournament exists and an (n,2q)-tournamentdoes not.If n = 2q, we can label the contestants and rounds by elements of the additivegroup Zq

2. If contestants x and x+ j meet in the round labeled j, it is easy toverify the conditions. If n = 2qp, we can divide the contestants into p disjointgroups of 2q and perform a (2q,2q" 1)-tournament in each, thus obtaining an(n,2q"1)-tournament.For the other direction let Gi be the graph of players with edges between any twoplayers who met in the first i rounds. We claim that the size of each connectedcomponent of Gi is a power of 2. For i = 1 this is obvious; assume that it holdsfor i. Suppose that the componentsC and D merge in the (i+1)th round. Then

746 4 Solutions

some c ( C and d ( D meet in this round. Moreover, each player in C meets aplayer in D. Indeed, for every c+ ( C there is a path c = c0,c1, . . . ,ck = c+ withc jc j+1 ( Gi; then if d j is the opponent of c j in the (i+1)th round, condition (ii)shows that each d jd j+1 belongs to Gi, so dk ( D. Analogously, all players in Dmeet players in C, so |C| = |D|, proving our claim. Now if there are 2q rounds,every component has size at least 2q+1 and is thus divisible by 2q+1, which isimpossible if 2q+1 ! n.

12. Let U and D be the sets of upward and downward unit triangles, respectively.Two triangles are neighbors if they form a diamond. For A3 D, denote by F(A)the set of neighbors of the elements of A.If a holey triangle can be tiled with diamonds, in every upward triangle of side lthere are l2 elements of D, so there must be at least as many elements of U andat most l holes.Now we pass to the other direction. It is enough to show the condition (ii) ofthe marriage theorem: For every set X 4 D we have |F(X)| ! |X |. Assume thecontrary, that |F(X)| < |X | for some set X . Note that any two elements of Dwith a common neighbor must share a vertex; this means that we can focuson connected sets X . Consider an upward triangle of side 3. It contains threeelements of D; if two of them are in X , adding the third one to X increases F(X)by at most 1, so |F(X)| < |X | still holds. Continuing this procedure, we will endup with a set X forming an upward subtriangle of T and satisfying |F(X)|< |X |,which contradicts the conditions of the problem. This contradiction proves that|F(X)|! |X | for every set X , and an application of the Hall’s marriage theoremestablishes the result.

13. Consider a polyhedron P with v vertices, e edges, and f faces. Consider themap ! to the unit sphere S taking each vertex, edge, or face x of P to the setof outward unit normal vectors (i.e., points on S) to the support planes of P

containing x. Thus ! maps faces to points on S, edges to shorter arcs of bigcircles connecting some pairs of these points, and vertices to spherical regionsformed by these arcs. These points, arcs, and regions on S form a “sphericalpolyhedron” G .We now translate the conditions of the problem into the language of G . Denoteby x the image of x through reflection with the center in the center of S. No edgeofP being parallel to another edge or face means that the big circle of any edgee of G does not contain any vertex V nonincident to e. Also note that vertices Aand B of P are antipodal if and only if !(A) and !(B) intersect, and that themidpoints of edges a and b are antipodal if and only if !(a) and !(b) intersect.Consider the unionF of G and G . The faces ofF are the intersections of facesof G and G , so their number equals 2A. Similarly, the edges of G and G have 2Bintersections, soF has 2e+4B edges and 2 f +2B vertices. NowEuler’s theoremforF gives us 2e+4B+2= 2A+2 f +2B, and therefore A"B= e" f +1.

14. The condition of the problem implies that !PBC+ !PCB = 90* "%/2, i.e.,!BPC = 90* +%/2 = !BIC. Thus P lies on the circumcircle # of .BCI. It is


well known that the center M of # is the second intersection of AI with the cir-cumcircle of.ABC. Therefore AP! AM"MP= AM"MI = AI, with equalityif and only if P% I.

15. The relation AK/KB = DL/LC implies that AD, BC, and KL have a commonpoint O. Moreover, since !APB= 180* "!ABC and !DQC = 180* "!BCD,line BC is tangent to the circles APB andCQD. These two circles are homotheticwith respect to O, so if OP meets circle APB again at P+, we have !PQC =!PP+B= !PBC, showing that P,Q,B,C lie on a circle.

16. Let the diagonals AC and BD meet at Q and AD and CE meet at R. The quadri-laterals ABCD and ACDE are similar, so AQ/QC = AR/RD. Now if AP meetsCD at M, Ceva’s theorem gives us CM

MD = CQQA ·

ARRD = 1.

17. Let M be the point on AC such that JM 0 KL. It is enough to prove that AM =2AL.From !BDA = % we obtain that !JDM = 90* " %

2 = !KLA = !JMD; henceJM = JD, and the tangency point of the incircle of.BCD with CD is the mid-point T of segmentMD. Therefore,DM = 2DT = BD+CD"BC= AB"BC+CD, which gives us

AM = AD+DM = AC+AB"BC= 2AL.

18. Assume that A1B1 and CJ intersectat K. Then JK is parallel and equalto C1D and DC1/C1J = JK/JB1 =JB1/JC = C1J/JC, so the right tri-angles DC1J and C1JC are similar;hence C1C 1 DJ. Thus E belongs toCC1. The points A1, B1, and E lie onthe circle with diameter CJ. Therefore!DBA1 = !A1CJ= !A1ED, implying

A

J

C1

B1

A1

B

C

D

K

E

that BEA1D is cyclic; hence !A1EB= 90*. Likewise, ADEB1 is cyclic because!EB1A= !EJC = !EDC1, so !AEB1 = 90*.Second solution. The segments JA1, JB1, JC1 are tangent to the circles withdiameters A1B, AB1, C1D. Since JA21 = JB21 = JC21 = JD · JE , E lies on the firsttwo circles (with diameters A1B and AB1), so !AEB1 = !A1EB= 90*.

19. The homothety with center E mapping#1 to # maps D to B, so D lies on BE;analogously,D lies on AF . Let AE andBF meet at point C. The lines BE andAF are altitudes of triangle ABC, so Dis the orthocenter and C lies on t. Letthe line through D parallel to AB meetAC atM. The centersO1 andO2 are themidpoints of DM and DN respectively.

A B

C

O K

E

FDO1M O2

P

748 4 Solutions

We have thus reduced the problem to a classical triangle geometry problem:If CD and EF intersect at P, we should prove that points A, O1 and P arecollinear (analogously, so are B, O2, P). By Menelaus’s theorem for triangleCDM, this is equivalent to CA

AM = CPPD , which is again equivalent to

CKKD = CP

PD(because DM 0 AB), where K is the foot of the altitude from C to AB. The lastequality immediately follows from the fact that the pairsC,D; P,K are harmoni-cally adjoint.

20. Let I be the incenter of .ABC. It is well known that TaTc and TaTb are theperpendicular bisectors of the segments BI and CI respectively. Let TaTb meetAC at P and #b at U , and let TaTc meet AB at Q and #c at V . We have !BIQ=!IBQ = !IBC, so IQ 0 BC; similarly IP 0 BC. Hence PQ is the line through Iparallel to BC.The homothety from Tb mapping #b to the circumcircle # of ABC maps thetangent t to #b at U to the tangent to # at Ta that is parallel to BC. It followsthat t 0 BC. Let t meet AC at X . Since XU = XMb and !PUMb = 90*, X is themidpoint of PMb. Similarly, the tangent to #c at V meets QMc at its midpoint Y .But since XY 0 PQ 0 MbMc, points U,X ,Y,V are collinear, so t coincides withthe common tangent pa. Thus pa runs midway between I andMbMc. Analogousconclusions hold for pb and pc, so these three lines form a triangle homotheticto the triangleMaMbMc from center I in ratio 1

2 , which is therefore similar to thetriangle ABC in ratio 1

4 .

21. The following proposition is easy to prove:Lemma. For an arbitrary point X inside a convex quadrilateral ABCD, the cir-

cumcircles of triangles ADX and BCX are tangent at X if and only if!ADX+!BCX = !AXB.

Let Q be the second intersection point of the circles ABP and CDP (we assumeQ &% P; the opposite case is similarly handled). It follows from the conditionsof the problem that Q lies inside quadrilateral ABCD (since !BCP+ !BAP <180*,C is outside the circumcircle of APB; the same holds for D). If Q is inside.APD (the other case is similar), we have !BQC= !BQP+!PQC= !BAP+!CDP ' 90*. Similarly, !AQD' 90*. Moreover, !ADQ+!BCQ= !ADP+!BCP = !APB = !AQB implies that circles ADQ and BCQ are tangent at Q.Therefore the interiors of the semicircles with diameters AD and BC are disjoint,and if M, N are the midpoints of AD and BC respectively, we have 2MN !AD+BC. On the other hand, 2MN ' AB+CD because "#BA+

"#CD = 2""#MN, and

the statement of the problem immediately follows.

22. We work with oriented angles modulo 180*. For two lines a,b we denote by!(l,m) the angle of counterclockwise rotation transforming a to b; also, by!ABC we mean !(BA,BC).It is well known that the circles AB1C1, BC1A1, andCA1B1 have a common point,say P. Let O be the circumcenter of ABC. Set !PB1C = !PC1A= !PA1B= * .Let A2P,B2P,C2P meet the circle ABC again at A4,B4,C4, respectively. Since


!A4A2A = !PA2A = !PC1A = ! and thus !A4OA = 2! etc., !ABC is theimage of!A4B4C4 under rotationR about O by 2! .Therefore !(AB4,PC1) = !B4AB +!AC1P = ! " ! = 0, so AB4 # PC1.Let PC1 intersect A4B4 at C5; defineA5,B5 analogously. Then !B4C5P =!A4B4A= ! , so AB4C5C1 is an isosce-les trapezoid with BC3 = AC1 = B4C5.Similarly, AC3 = A4C5, soC3 is the im-age of C5 under R; similar statementshold for A5,B5. Thus !A3B3C3 $=!A5B5C5. It remains to show that!A5B5C5 $!A2B2C2.We have seen that!A4B5P= !B4C5P,

A B

C

P

A1

B1C1

A2

B2

C2

A4

B4

C4

A5 B5

C5

which implies that P lies on the circle A4B5C5. Analogously, P lies on the circleC4A5B5. Therefore

!A2B2C2 = !A2B2B4+!B4B2C2 = !A2A4B4+!B4C4C2= !PA4C5+!A5C4P= !PB5C5+!A5B5P= !A5B5C5,

and similarly for the other angles, which is what we wanted.

23. Let Si be the area assigned to side AiAi+1 of polygon P = A1 . . .An of area S.We start with the following auxiliary statement.Lemma. At least one of the areas S1, . . . ,Sn is not smaller than 2S/n.Proof. It suffices to prove the statement for even n. The case of odd n will

then follow immediately from this case applied to the degenerate 2n-gonA1A%1 . . .AnA%n, where A%i is the midpoint of AiAi+1.Let n = 2m. For i = 1,2, . . . ,m, denote by Ti the area of the region Piinside the polygon bounded by the diagonals AiAm+i, Ai+1Am+i+1 and thesides AiAi+1, Am+iAm+i+1. We observe that the regionsPi cover the entirepolygon. Indeed, let X be an arbitrary point inside the polygon, to the left(without loss of generality) of the ray A1Am+1.Then X is to the right of the rayAm+1A1, so there is a k such that X isto the left of ray AkAk+m and to theright of ray Ak+1Ak+m+1, i.e., X &Pk. It follows that T1+ · · ·+Tm' S;hence at least one Ti is not smallerthan 2S/n, say T1 ' 2S/n.Let O be the intersection point ofA1Am+1 and A2Am+2, and let usassume without loss of generality

A1 A2A3

Am+1Am+2Am+3

OX

P1

P1

that SA1A2O ' SAm+1Am+2O and A1O ' OAm+1. Then required result nowfollows from

S1 ' SA1A2Am+2 = SA1A2O+SA1Am+2O ' SA1A2O+SAm+1Am+2O = T1 '2Sn

.

750 4 Solutions

If, contrary to the assertion, $ SiS < 2, we can choose rational numbers qi =

2mi/N with N = m1 + · · ·+mn such that qi > Si/S. However, considering thegiven polygon as a degenerate N-gon obtained by division of side AiAi+1 intomi equal parts for each i and applying the lemma, we obtain Si/mi ! 2S/N, i.e.,Si/S! qi for some i, a contradiction.Equality holds if and only ifP is centrally symmetric.Second solution. We say that vertexV is assigned to side a of a convex (possiblydegenerate) polygonP if the triangle determined by a and V has the maximumarea Sa among the triangles with side a contained in P . Define !(P) = $a Saand ( (P) = !(P)"2[P]. We use induction on the number n of pairwise non-parallel sides of P to show that ( (P) ! 0 for every polygonP . This is obvi-ously true for n= 2, so let n! 3.There exist two adjacent sides AB and BC whose respective assigned verticesU and V are distinct. Let the lines through U and V parallel to AB and BC re-spectively intersect at point X . Assume, without loss of generality, that there areno sides of P lying on UX and VX . Call the sides and vertices of P lyingwithin the triangle UVX passive (excluding vertices U and V ). It is easy to seethat no passive vertex is assigned to any side ofP and that vertex B is assignedto every passive side. Now replace all passive vertices of P by X , obtaining apolygonP +. Vertex B is assigned to sidesUX andVX ofP +. Therefore the sumof areas assigned to passive sides increases by the area S of the part of quadri-lateral BUXV lying outsideP; the other assigned areas do not change. Thus !increases by S. On the other hand, the area of the polygon also increases by S,so ( must decrease by S.Note that the change from P to P + decreases the number of nonparallel sides.Thus by the inductive hypothesis we have ( (P) ! ( (P +) ! 0.Third solution. To each convex n-gon P = A1A2 . . .An we assign a centrallysymmetric 2n-gonQ, called the associate of P , as follows. Attach the 2n vec-tors ±""""#

AiAi+1 at a common origin and label them b1, . . . , b2n counterclockwiseso that bn+i = "bi for 1 ' i ' n. Then take Q to be the polygon B1B2 . . .B2nwith """"#BiBi+1 = bi. Denote by ai the side ofP corresponding to bi (i= 1, . . . ,n).The distance between the parallel sides BiBi+1 and Bn+iBn+i+1 ofQ equals twicethe maximum height ofP to the side ai. Thus, ifO is the center ofQ, the area of.BiBi+1O (i= 1, . . . ,n) is exactly the area Si assigned to side ai ofP; therefore[Q] = 2$Si. It remains to show that d(P) = [Q]"4[P]! 0.(i) Suppose thatP has two parallel sides ai and a j, where a j ! ai, and removefrom it the parallelogram D determined by ai and a part of side a j. Weobtain a polygonP + with a smaller number of nonparallel sides. Then theassociate of P + is obtained from Q by removing a parallelogram similarto D in ratio 2 (and with area four times that of D); thus d(P +) = d(P).

(ii) Suppose that there is a side bi (i' n) ofQ such that the sum of the anglesat its endpoints is greater than 180*. Extend the pairs of sides adjacent to biand bn+i to their intersectionsU and V , thus enlargingQ by two congruenttriangles to a polygon Q+. Then Q+ is the associate of the polygon P +


obtained from P by attaching a triangle congruent to BiBi+1U to the sideai. Therefore d(P +) equals d(P) minus twice the area of the attachedtriangle.

By repeatedly performing the operations (i) and (ii) to polygonP we will even-tually reduce it to a parallelogram E , thereby decreasing the value of d. Sinced(E) = 0, it follows that d(P) ! 0.Remark. PolygonQ is the Minkowski sum ofP and a polygon centrally sym-metric to P . Thus the inequality [Q] ! 4[P] is a direct consequence of theBrunn–Minkowski inequality.

24. Obviously x ! 0. For x = 0 the only solutions are (0,±2). Now let (x,y) be asolution with x> 0. Without loss of generality, assume that y> 0. The equationrewritten as 2x(1+2x+1) = (y"1)(y+1) shows that one of the factors y±1 isdivisible by 2 but not by 4 and the other by 2x"1 but not by 2x; hence x ! 3.Thus y = 2x"1m+ ) , where m is odd and ) = ±1. Plugging this in the originalequation and simplifying yields

2x"2(m2"8) = 1" )m. (1)

Since m= 1 is obviously impossible, we havem! 3 and hence ) ="1. Now (1)gives us 2(m2" 8) ' 1+m, implying m = 3, which leads to x = 4 and y = 23.Thus all solutions are (0,±2) and (4,±23).

25. If x is rational, its digits repeat periodically starting at some point. If n is thelength of the period of x, the sequence 2,22,23, . . . is eventually periodic modulon, so the corresponding digits of x (i.e., the digits of y) also make an eventuallyperiodic sequence, implying that y is rational.

26. Consider g(n) = [ n1 ]+ [ n2 ]+ · · ·+[ nn ] = n f (n) and define g(0) = 0. Since for anyk the difference [ nk ]" [ n"1k ] equals 1 if k divides n and 0 otherwise, we obtainthat g(n) = g(n" 1) + d(n), where d(n) is the number of positive divisors ofn. Thus g(n) = d(1)+ d(2)+ · · ·+ d(n) and f (n) is the arithmetic mean of thenumbers d(1), . . . ,d(n). Therefore, (a) and (b) will follow if we show that each ofd(n+1) > f (n) and d(n+1) < f (n) holds infinitely often. But d(n+1) < f (n)holds whenever n+1 is prime, and d(n+1) > f (n) holds whenever d(n+1) >d(1), . . . ,d(n) (which clearly holds for infinitely many n).

27. We first show that every fixed point x of Q is in fact a fixed point of P * P.Consider the sequence given by x0 = x and xi+1 = P(xi) for i! 0. Assume xk =x0. We know that u" v divides P(u)"P(v) for every two distinct integers u andv. In particular,

di = xi+1" xi | P(xi+1)"P(xi) = xi+2" xi+1 = di+1

for all i, which together with dk = d0 implies |d0| = |d1| = · · · = |dk|. Supposethat d1 = d0 = d &= 0. Then d2 = d (otherwise x3 = x1 and x0 will never occur inthe sequence again). Similarly, d3 = d etc., and hence xi = x0+ id &= x0 for alli, a contradiction. It follows that d1 = "d0, so x2 = x0 as claimed. Thus we canassume that Q= P*P.

752 4 Solutions

If every integer t with P(P(t)) = t also satisfies P(t) = t, the number of solutionsis clearly at most degP = n. Suppose that P(t1) = t2, P(t2) = t1, P(t3) = t4,and P(t4) = t3, where t1 &= t2,3,4 (but not necessarily t3 &= t4). Since t1" t3 dividest2" t4 and vice versa, we conclude that t1" t3 =±(t2" t4). Assume that t1" t3 =t2"t4, i.e. t1"t2 = t3"t4 = u &= 0. Since the relation t1"t4 =±(t2"t3) similarlyholds, we obtain t1 " t3 + u = ±(t1" t3 " u) which is impossible. Therefore,we must have t1" t3 = t4" t2, which gives us P(t1) + t1 = P(t3) + t3 = c forsome c. It follows that all integral solutions t of the equation P(P(t)) = t satisfyP(t)+ t = c, and hence their number does not exceed n.

28. Every prime divisor p of x7"1x"1 = x6+ · · ·+ x+ 1 is congruent to 0 or 1 modulo7. Indeed, if p | x"1, then x7"1

x"1 % 1+ · · ·+1% 7 (mod p), so p = 7; otherwisethe order of x modulo p is 7 and hence p % 1 (mod 7). Therefore every positivedivisor d of x7"1x"1 satisfies d % 0 or 1 (mod 7).Now suppose (x,y) is a solution of the given equation. Since y"1 and y4+ y3+y2+ y+1 divide x7"1

x"1 = y5"1, we have y% 1 or 2 and y4+ y3+ y2+ y+1% 0or 1 (mod 7). However, y% 1 or 2 implies that y4+y3+y2+y+1% 5 or 3 (mod7), which is impossible.

29. All representations of n in the form ax+ by (x,y ( Z) are given by (x,y) =(x0 + bt,y0" at), where x0,y0 are fixed and t ( Z is arbitrary. The followinglemma enables us to determine w(n).Lemma. The equality w(ax+ by) = |x|+ |y| holds if and only if one of the

following conditions holds:(i) a"b

2 < y' a+b2 and x! y" a+b

2 ;(ii) " a"b

2 ' y' a"b2 and x ( Z;

(iii) " a+b2 ' y< " a"b

2 and x' y+ a+b2 .

Proof. Without loss of generality, assume that y ! 0. We have w(ax+ by) =|x|+y if and only if |x+b|+ |y"a|! |x|+y and |x"b|+(y+a)! |x|+y,where the latter is obviously true and the former clearly implies y< a. Thenthe former inequality becomes |x+b|" |x|! 2y"a. We distinguish threecases: if y' a"b

2 , then 2y"a' b and the previous inequality always holds;for a"b2 < y ' a+b

2 , it holds if and only if x ! y" a+b2 ; and for y > a+b

2 , itnever holds.

Now let n= ax+by be a local champion with w(n) = |x|+ |y|. As in the lemma,we distinguish three cases:(i) a"b

2 < y' a+b2 . Then x+1! y" a+b

2 by the lemma, sow(n+a)= |x+1|+y(because n+a= a(x+1)+by). Sincew(n+a)'w(n), we must have x< 0.Likewise, w(n"a) equals either |x"1|+y=w(n)+1 or |x+b"1|+a"y.The condition w(n"a)' w(n) leads to x' y" a+b"1

2 ; hence x= y" [ a+b2 ]

and w(n) = [ a+b2 ]. Now w(n"b) = "x+ y"1= w(n)"1 and w(n+b) =

(x+b)+ (a"1" y)= a+b"1" [ a+b2 ] ' w(n), so n is a local champion.Conversely, every n = ax+ by with a"b

2 < y ' a+b2 and x = y" [ a+b2 ] is


a local champion. Thus we obtain b" 1 local champions, which are alldistinct.

(ii) |y| ' a"b2 . Now we conclude from the lemma that w(n"a) = |x"1|+ |y|

and w(n+ a) = |x+ 1|+ |y|, and at least one of these two values exceedsw(n) = |x|+ |y|. Thus n is not a local champion.

(iii) " a+b2 ' y< " a"b

2 . By taking x,y to "x,"y this case is reduced to case (i),so we again have b"1 local champions n= ax+by with x= y+[ a+b2 ].

It is easy to check that the sets of local champions from cases (i) and (iii) coincideif a and b are both odd (so we have b" 1 local champions in total), and areotherwise disjoint (then we have 2(b"1) local champions).

30. We shall show by induction on n that there exists an arbitrarily largem satisfying2m %"m (mod n). The case n= 1 is trivial; assume that n> 1.Recall that the sequence of powers of 2 modulo n is eventually periodic with theperiod dividing *(n); thus 2x % 2y whenever x % y (mod *(n)) and x and y arelarge enough. Let us consider m of the form m % "2k (mod n*(n)). Then thecongruence 2m % "m (mod n) is equivalent to 2m % 2k (mod n), and this holdswhenever"2k %m% k (mod *(n)) and m,k are large enough. But the existenceof m and k is guartanteed by the inductive hypothesis for *(n), so the inductionis complete.

A

Notation and Abbreviations

A.1 Notation

We assume familiarity with standard elementary notation of set theory, algebra, logic,geometry (including vectors), analysis, number theory (including divisibility andcongruences), and combinatorics. We use this notation liberally.We assume familiarity with the basic elements of the game of chess (the movementof pieces and the coloring of the board).The following is notation that deserves additional clarification.

* B(A,B,C), A"B"C: indicates the relation of betweenness, i.e., that B is be-tween A and C (this automatically means that A,B,C are different collinearpoints).

* A= l16 l2: indicates that A is the intersection point of the lines l1 and l2.

* AB: line through A and B, segment AB, length of segment AB (depending oncontext).

* [AB: ray starting in A and containing B.

* (AB: ray starting in A and containing B, but without the point A.

* (AB): open interval AB, set of points between A and B.

* [AB]: closed interval AB, segment AB, (AB)7{A,B}.

* (AB]: semiopen interval AB, closed at B and open at A, (AB)7{B}.The same bracket notation is applied to real numbers, e.g., [a,b) = {x | a' x<b}.

* ABC: plane determined by points A,B,C, triangle ABC (.ABC) (depending oncontext).

* [AB,C: half-plane consisting of line AB and all points in the plane on the sameside of AB as C.

* (AB,C: [AB,C without the line AB.

792 A Notation and Abbreviations

* 8"#a ,"#b 9, "#a ·

"#b : scalar product of "#a and

"#b .

* a,b,c,%,& ,': the respective sides and angles of triangle ABC (unless otherwiseindicated).

* k(O,r): circle k with center O and radius r.

* d(A, p): distance from point A to line p.

* SA1A2...An , [A1A2 . . .An]: area of n-gon A1A2 . . .An (special case for n = 3, SABC:area of.ABC).

* N, Z, Q, R, C: the sets of natural, integer, rational, real, complex numbers (re-spectively).

* Zn: the ring of residues modulo n, n ( N.

* Zp: the field of residues modulo p, p being prime.

* Z[x], R[x]: the rings of polynomials in x with integer and real coefficients re-spectively.

* R:: the set of nonzero elements of a ring R.

* R[%], R(%), where % is a root of a quadratic polynomial in R[x]: {a+b% | a,b (R}.

* X0: X 7{0} for X such that 0 /( X .

* X+, X", aX+b, aX+bY : {x | x ( X ,x> 0}, {x | x ( X ,x< 0}, {ax+b | x ( X},{ax+by | x ( X ,y (Y} (respectively) for X ,Y 3 R, a,b ( R.

* [x], ;x<: the greatest integer smaller than or equal to x.

* ,x-: the smallest integer greater than or equal to x.

The following is notation simultaneously used in different concepts (depending oncontext).

* |AB|, |x|, |S|: the distance between two points AB, the absolute value of the num-ber x, the number of elements of the set S (respectively).

* (x,y), (m,n), (a,b): (ordered) pair x and y, the greatest common divisor of inte-gers m and n, the open interval between real numbers a and b (respectively).

A.2 Abbreviations

We tried to avoid using nonstandard notation and abbreviations as much as possible.However, one nonstandard abbreviation stood out as particularly convenient:

* w.l.o.g.: without loss of generality.

Other abbreviations include:

* RHS: right-hand side (of a given equation).

A.2 Abbreviations 793

* LHS: left-hand side (of a given equation).

* QM, AM, GM, HM: the quadratic mean, the arithmetic mean, the geometricmean, the harmonic mean (respectively).

* gcd, lcm: greatest common divisor, least common multiple (respectively).

* i.e.: in other words.

* e.g.: for example.

B

Codes of the Countries of Origin

ARG ArgentinaARM ArmeniaAUS AustraliaAUT AustriaBEL BelgiumBGR BulgariaBLR BelarusBRA BrazilCAN CanadaCHN ChinaCOL ColombiaCUB CubaCYP CyprusCZE Czech RepublicCZS CzechoslovakiaESP SpainEST EstoniaFIN FinlandFRA FranceFRG Germany, FRGDR Germany, DRGEO GeorgiaGER GermanyHEL GreeceHKG Hong Kong

HRV CroatiaHUN HungaryIDN IndonesiaIND IndiaIRL IrelandIRN IranISL IcelandISR IsraelITA ItalyJPN JapanKAZ KazakhstanKOR Korea, SouthKWT KuwaitLTU LithuaniaLUX LuxembourgLVA LatviaMAR MoroccoMEX MexicoMKD MacedoniaMNG MongoliaNLD NetherlandsNOR NorwayNZL New ZealandPER PeruPHI Philippines

POL PolandPOR PortugalPRI Puerto RicoPRK Korea, NorthROU RomaniaRUS RussiaSAF South AfricaSCG Serbia and

MontenegroSGP SingaporeSRB SerbiaSVK SlovakiaSVN SloveniaSWE SwedenTHA ThailandTUN TunisiaTUR TurkeyTWN TaiwanUKR UkraineUNK United KingdomUSA United StatesUSS Soviet UnionUZB UzbekistanVNM VietnamYUG Yugoslavia

C

Authors of Problems

1959-02 C. Ionescu-Tiu, ROU – IMO21959-06 Cezar Cosnita, ROU – IMO6

1960-03 G.D. Simionescu, ROU – IMO3


1962-04 Cezar Cosnita, ROU – IMO4

1963-05 Wolfgang Engel, GDR – IMO5



1967-04 Tullio Viola, ITA – IMO4

1968-06 David Monk, UNK – IMO6

1969-01 Wolfgang Engel, GDR – IMO11969-05 Abish Mekei, MNG – IMO5

1970-02 D. Batinetzu, ROU – IMO21970-06 D. Gerll, A. Warusfel, FRA1970-10 Åke Samuelsson, SWE – IMO6

1973-05 Georges Glaeser, FRA1973-10 Åke Samuelsson, SWE – IMO61973-11 Åke Samuelsson, SWE – IMO31973-13 !orde Dugo"ija, YUG1973-14 !orde Dugo"ija, YUG – IMO4

1974-01 Murray Klamkin, USA – IMO11974-06 Ciprian Borcea, ROU – IMO31974-08 Jan van de Craats, NLD – IMO51974-10 Matti Lehtinen, FIN – IMO2

1975-08 Jan van de Craats, NLD – IMO31975-10 David Monk, UNK – IMO61975-14 Vladimir Jankovic, YUG

1975-10 R. Lyness,D. Monk, UNK – IMO6

1976-05 Jan van de Craats, NLD – IMO51976-06 Jan van de Craats, NLD – IMO31976-09 Matti Lehtinen, FIN – IMO21976-10 Murray Klamkin, USA – IMO4

1977-01 Ivan Prodanov, BGR – IMO61977-03 Hermann Frasch, FRG – IMO51977-04 Arthur Engel, FRG1977-05 Arthur Engel, FRG1977-06 Helmut Bausch, GDR1977-07 David Monk, UNK – IMO41977-10 Jan van de Craats, NLD – IMO31977-12 Jan van de Craats, NLD – IMO11977-15 P. !"c Chính, VNM – IMO2

1978-10 Jan van de Craats, NLD – IMO61978-12 Murray Klamkin, USA – IMO41978-13 Murray Klamkin, USA – IMO21978-15 Dragoslav Ljubic, YUG1978-16 Dragoslav Ljubic, YUG

1979-04 N. Hadzhiivanov,N. Nenov, BGR – IMO2

1979-07 Arthur Engel, FRG – IMO11979-08 Arthur Engel, FRG1979-09 Arthur Engel, FRG – IMO61979-12 Hans-Dietrich Gronau, GDR

798 C Authors of Problems

1979-15 Joe Gillis, ISR – IMO51979-22 N. Vasilyev,

I.F. Sharygin, USS – IMO31979-25 Murray Klamkin, USA – IMO41979-26 Milan Bo#ic, YUG

1981-05 Rafael Marino, COL1981-07 Juha Oikkonen, FIN – IMO61981-08 Arthur Engel, FRG – IMO21981-09 Arthur Engel, FRG1981-12 Jan van de Craats, NLD – IMO31981-15 David Monk, UNK – IMO11981-19 Vladimir Jankovic, YUG

1981-01 David Monk, UNK – IMO11982-03 A.N. Grishkov, USS – IMO31982-05 Jan van de Craats, NLD – IMO51982-06 V. Nh# C#$ng, VNM – IMO61982-13 Jan van de Craats, NLD – IMO21982-16 David Monk, UNK – IMO4

1983-08 Juan Ochoa, ESP1983-09 Murray Klamkin, USA – IMO61983-12 David Monk, UNK – IMO11983-13 Lucien Kieffer, LUX1983-14 Marcin Kuczma, POL – IMO51983-16 Konrad Engel, GDR1983-17 Hans-Dietrich Gronau, GDR1983-18 Arthur Engel, FRG – IMO31983-19 Titu Andreescu, ROU1983-23 Igor F. Sharygin, USS – IMO2

1984-04 T. Dashdorj, MNG – IMO51984-05 M. Stoll,

B. Haible, FRG – IMO11984-07 Horst Sewerin, FRG1984-08 Ioan Tomescu, ROU – IMO31984-12 Aiko Tiggelaar, NLD – IMO21984-14 L. Panaitopol, ROU – IMO41984-15 Lucien Kieffer, LUX1984-17 Arthur Engel, FRG

1985-01 R. Gonchigdorj, MNG – IMO41985-03 Jan van de Craats, NLD – IMO31985-04 George Szekeres, AUS – IMO21985-17 Åke Samuelsson, SWE – IMO6

1985-20 Frank Budden, UNK – IMO11985-22 Igor F. Sharygin, USS – IMO5

1986-01 David Monk, UNK – IMO51986-05 Arthur Engel, FRG – IMO11986-06 Johannes Notenboom, NLD1986-08 Cecil Rousseau, USA1986-09 Konrad Engel, GDR – IMO61986-12 Elias Wegert, GDR – IMO31986-13 Arthur Engel, FRG1986-15 Johannes Notenboom, NLD1986-16 Sven Sigur$sson, ISL – IMO41986-17 G. Chang, D. Qi, CHN – IMO2

1987-06 Dimitris Kontogiannis, HEL1987-13 H.-D. Gronau, GDR – IMO51987-14 Arthur Engel, FRG1987-15 Arthur Engel, FRG – IMO31987-16 Horst Sewerin, FRG – IMO11987-20 Vsevolod F. Lev, USS – IMO61987-21 I.A. Kushnir, USS – IMO21987-22 N.Minh !"c, VNM – IMO4

1988-09 Stephan Beck, FRG – IMO61988-12 Dimitris Kontogiannis, HEL1988-13 D. Kontogiannis, HEL – IMO51988-16 Finbarr Holland, IRL – IMO41988-18 Lucien Kieffer, LUX – IMO11988-26 David Monk, UNK – IMO3

1989-01 Geoffrey Bailey, AUS – IMO21989-05 Joaquín Valderrama, COL1989-10 Theodoros Bolis, HEL1989-13 Eggert Briem, ISL – IMO41989-14 S.A. Shirali, IND1989-15 Fergus Gaines, IRL1989-20 Harm Derksen, NLD – IMO31989-21 Johannes Notenboom, NLD1989-22 Jose Marasigan, PHI – IMO11989-23 Marcin Kuczma, POL – IMO61989-25 Myung-Hwan Kim, KOR1989-30 Bernt Lindström, SWE – IMO5

1990-03 Pavol Cernek, CZS – IMO21990-06Hagen von Eitzen, FRG – IMO51990-07 Theodoros Bolis, HEL1990-11 C.R. Pranesachar, IND – IMO1

C Authors of Problems 799

1990-12 Fergus Gaines, IRL1990-14 Toshio Seimiya, JPN1990-16 Harm Derksen, NLD – IMO61990-17 Johannes Notenboom, NLD1990-18 Istvan Beck, NOR1990-23 L. Panaitopol, ROU – IMO31990-25 Albert Erkip, TUR – IMO4

1991-02 Toshio Seimiya, JPN1991-04 Johan Yebbou, FRA – IMO51991-06 A. Skopenkov, USS – IMO11991-08 Harm Derksen, NLD1991-10 Cecil Rousseau, USA – IMO41991-12 Chengzhang Li, CHN – IMO31991-16 L. Panaitopol, ROU – IMO21991-19 Tom Laffey, IRL1991-20 Tom Laffey, IRL1991-23 C.R. Pranesachar, IND1991-24 C.R. Pranesachar, IND1991-28 Harm Derksen, NLD – IMO6

1992-04 Chengzhang Li, CHN – IMO31992-05 Germán Rincón, COL1992-06 B.J. Venkatachala, IND – IMO21992-07 S.A. Shirali, IND1992-08 C.R. Pranesachar, IND1992-10 Stefano Mortola, ITA – IMO51992-13 Achim Zulauf, NZL – IMO11992-14 Marcin Kuczma, POL1992-18 Robin Pemantle, USA1992-19 Tom Laffey, IRL1992-20 Johan Yebbou, FRA – IMO41992-21 Tony Gardiner, UNK – IMO6

1993-03 Francisco Bellot Rosado, ESP1993-04 Francisco Bellot Rosado, ESP1993-05 Kerkko Luosto, FIN – IMO31993-06 Elias Wegert, GER – IMO51993-08 S.A. Shirali, IND1993-09 V.S. Joshi, IND1993-10 C.S. Yogananda, IND1993-11 Tom Laffey, IRL – IMO11993-12 Fergus Gaines, IRL1993-13 Tom Laffey, IRL1993-15 D. Dimovski, MKD – IMO41993-17 N.G. de Bruijn, NLD – IMO6

1993-21 Chirstopher Bradley, UNK1993-22 David Monk, UNK – IMO21993-23 David Monk, UNK1993-24 Titu Andreescu, USA

1994-01 Titu Andreescu, USA1994-02 V. Lafforgue, FRA – IMO11994-03 David Monk, UNK – IMO51994-05 Marcin Kuczma, POL1994-06 Vadym Radchenko, UKR1994-07 David Berenstein, COL1994-09 H. Nestra, R. Palm, EST1994-14 Vyacheslav Yasinskiy, UKR1994-16 H. Lausch,

G. Tonoyan, AUS/ARM – IMO21994-19 Hans Lausch, AUS – IMO41994-20 Kerkko Luosto, FIN – IMO61994-22 Gabriel Istrate, ROU – IMO31994-24 David Monk, UNK

1995-01Nazar Agakhanov,RUS – IMO21995-03 Vadym Radchenko, UKR1995-04 Titu Andreescu, USA1995-05 Vadym Radchenko, UKR1995-06 Toru Yasuda, JPN1995-07 B. Mihailov, BGR – IMO11995-08 Arthur Engel, GER1995-10 Vyacheslav Yasinskiy, UKR1995-11 A. McNaughton, NZL – IMO51995-12 Titu Andreescu, USA1995-14 Germán Rincón, COL1995-16 Alexander S. Golovanov, RUS1995-17 Jaromír %im"a, CZE – IMO31995-19 Tom Laffey, IRL1995-20 Marcin Kuczma, POL – IMO61995-22 Stephan Beck, GER1995-23 Igor Mitelman, UKR1995-24 Marcin Kuczma, POL – IMO41995-25 Marcin Kuczma, POL1995-28 B.J. Venkatachala, IND

1996-01 Peter Anastasov, SVN1996-02 F. Gaines, T. Laffey, IRL1996-03 Michalis Lambrou, HEL1996-06 Tom Laffey, IRL1996-08 M. Becheanu, ROU – IMO3


1996-09 Marcin Kuczma, POL1996-10 David Monk, UNK1996-11 J.P. Grossman, CAN – IMO21996-12 David Monk, UNK1996-13 Titu Andreescu, USA1996-14 N.M. Sedrakyan, ARM – IMO51996-16 Christopher Bradley, UNK1996-18 Vyacheslav Yasinskiy, UKR1996-19 Igor Mitelman, UKR1996-20 A.S. Golovanov, RUS – IMO41996-21 Emil Kolev, BGR1996-24 Kerkko Luosto, FIN – IMO11996-25 Vadym Radchenko, UKR1996-26 T. Andreescu, K. Kedlaya, USA1996-28 Laurent Rosaz, FRA – IMO61996-30 F. Gaines, T. Laffey, IRL

1997-01 Igor Voronovich, BLR – IMO11997-03 Arthur Engel, GER1997-04 E.S. Mahmoodian,

M. Mahdian, IRN – IMO41997-05 Mircea Becheanu, ROU1997-06 F. Gaines, T. Laffey, IRL1997-07 Valentina Kirichenko, RUS1997-08 David Monk, UNK – IMO21997-09 T. Andreescu, K. Kedlaya, USA1997-11 N.G. de Bruijn, NLD1997-12 Roberto Dvornicich, ITA1997-13 C.R. Pranesachar, IND1997-14 B.J. Venkatachala, IND1997-15 Alexander S. Golovanov, RUS1997-16 Igor Voronovich, BLR1997-17 Petr Kanovsk&, CZE – IMO51997-18 David Monk, UNK1997-19 Finbarr Holland, IRL1997-20 Kevin Hutchinson, IRL1997-21 A. Kachurovskiy,

O.Bogopolskiy, RUS – IMO31997-22 Vadym Radchenko, UKR1997-23 Christopher Bradley, UNK1997-24 G. Alkauskas, LTU – IMO61997-25 Marcin Kuczma, POL1997-26 Stefano Mortola, ITA

1998-01 Charles Leytem, LUX – IMO11998-02 Waldemar Pompe, POL

1998-03 V. Yasinskiy, UKR – IMO51998-06 Waldemar Pompe, POL1998-07 David Monk, UNK1998-08 Sambuddha Roy, IND1998-12 Marcin Kuczma, POL1998-13 O. Mushkarov,

N. Nikolov, BGR – IMO61998-14 David Monk, UNK – IMO41998-16 Vadym Radchenko, UKR1998-17 David Monk, UNK1998-19 Igor Voronovich, BLR – IMO31998-21 Murray Klamkin, CAN1998-22 Vadym Radchenko, UKR1998-25 Arkadii Slinko, NZL1998-26 Ravi B. Bapat, IND – IMO2

1999-01 Liang-Ju Chu, TWN – IMO41999-02 Nairi M. Sedrakyan, ARM1999-07 Nairi M. Sedrakyan, ARM1999-08 Shin Hitotsumatsu, JPN1999-09 Jan Villemson, EST – IMO11999-10 David Monk, UNK1999-12 P. Kozhevnikov, RUS – IMO51999-15 Marcin Kuczma, POL – IMO21999-19 Tetsuya Ando, JPN – IMO61999-21 C.R. Pranesachar, IND1999-22 Andy Liu, jury, CAN1999-23 David Monk, UNK1999-24 Ben Green, UNK1999-25 Ye. Barabanov,

I.Voronovich, BLR – IMO31999-26 Mansur Boase, UNK

2000-01 Sándor Dobos, HUN – IMO42000-02 Roberto Dvornicich, ITA2000-03 Federico Ardila, COL2000-07 Titu Andreescu, USA – IMO22000-08 I. Leader, P. Shiu, UNK2000-10 David Monk, UNK2000-12 Gordon Lessells, IRL2000-16 Valeriy Senderov, RUS – IMO52000-21 Sergey Berlov, RUS – IMO12000-22 C.R. Pranesachar, IND2000-24 David Monk, UNK2000-27 L. Emelyanov,

T. Emelyanova, RUS – IMO6


2001-01 B. Rajarama Bhat, IND2001-02 Marcin Kuczma, POL2001-04 Juozas Juvencijus Macys, LTU2001-05 O. Mushkarov,

N. Nikolov, BGR2001-06 Hojoo Lee, KOR – IMO22001-07 Federico Ardila, COL2001-08 Bill Sands, CAN – IMO42001-10 Michael Albert, NZL2001-12 Bill Sands, CAN2001-14 Christian Bey, GER – IMO32001-15 Vyacheslav Yasinskiy, UKR2001-16 Hojoo Lee, KOR – IMO12001-17 Christopher Bradley, UNK2001-19 Sotiris Louridas, HEL2001-20 C.R. Pranesachar, IND2001-22 Shay Gueron, ISR – IMO52001-23 Australian PSC, AUS2001-24 Sandor Ortegón, COL2001-25 Kevin Buzzard, UNK2001-27 A. Ivanov, BGR – IMO62001-28 F. Petrov, D. !ukic, RUS

2002-02 Mihai Manea, ROU – IMO42002-03 G. Bayarmagnai, MNG2002-04 Stephan Beck, GER2002-06 L. Panaitopol, ROU – IMO32002-08 Hojoo Lee, KOR2002-09 Hojoo Lee, KOR – IMO22002-11 Angelo Di Pasquale, AUS2002-12 V. Yasinskiy, UKR – IMO62002-16 Du"an !ukic, SCG2002-17 Marcin Kuczma, POL2002-18 B.J. Venkatachala, IND – IMO52002-19 C.R. Pranesachar, IND2002-20 Omid Naghshineh, IRN2002-21 Federico Ardila, COL – IMO12002-23 Federico Ardila, COL2002-24 Emil Kolev, BGR2002-26 Marcin Kuczma, POL2002-27 Michael Albert, NZL

2003-01 Kiran Kedlaya, USA2003-02 A. Di Pasquale,

D. Mathews, AUS2003-04 Finbarr Holland, IRL – IMO5

2003-05 Hojoo Lee, KOR2003-06 Reid Barton, USA2003-07 C.G. Moreira, BRA – IMO12003-09 Juozas Juvencijus Macys, LTU2003-12 Dirk Laurie, SAF2003-13 Matti Lehtinen, FIN – IMO42003-15 C.R. Pranesachar, IND2003-17 Hojoo Lee, KOR2003-18Waldemar Pompe, POL – IMO32003-19 Dirk Laurie, SAF2003-20 Marcin Kuczma, POL2003-21 Zoran %unic, USA2003-22 A. Ivanov, BGR – IMO22003-23 Laurentiu Panaitopol, ROU2003-24 Hojoo Lee, KOR2003-25 Johan Yebbou, FRA – IMO6

2004-01 Hojoo Lee, KOR – IMO42004-02 Mihai Baluna, ROU2004-03 Dan Brown, CAN2004-04 Hojoo Lee, KOR – IMO22004-07 Finbarr Holland, IRL2004-08 Guihua Gong, PRI2004-09 Horst Sewerin, GER2004-10 Norman Do, AUS2004-11 Marcin Kuczma, POL2004-12 A. Slinko, S. Marshall, NZL2004-14 J. Villemson,

M. Pettai, EST – IMO32004-15 Marcin Kuczma, POL2004-16 D. Serbanescu,

V. Vornicu, ROU – IMO12004-18 Hojoo Lee, KOR2004-19Waldemar Pompe, POL – IMO52004-20 Du"an !ukic, SCG2004-21 B. Green, E. Crane, UNK2004-23 Du"an !ukic, SCG2004-26 Mohsen Jamali, IRN2004-27 Jaros'aw Wróblewski, POL2004-28 M. Jamali,

A. Morabi, IRN – IMO62004-29 John Murray, IRL2004-30 Alexander Ivanov, BGR

2005-02 Nikolai Nikolov, BGR2005-04 B.J. Venkatachala, IND


2005-05 Hojoo Lee, KOR – IMO32005-06 Australian PSC, AUS2005-09 Federico Ardila, COL2005-10 Du"an !ukic, SCG2005-11 R. Gologan,

D. Schwarz, ROU – IMO62005-12 R. Liu, Z. Feng, USA2005-13 Alexander Ivanov, BGR2005-14 Dimitris Kontogiannis, HEL2005-15 Bogdan Enescu, ROU – IMO12005-16 Vyacheslav Yasinskiy, UKR2005-17Waldemar Pompe, POL – IMO52005-20 Hojoo Lee, KOR2005-21 Mariusz Ska'ba, POL – IMO42005-22 N.G. de Bruijn, NLD – IMO22005-24 Carlos Caicedo, COL2005-26 Mohsen Jamali, IRN

2006-01 Härmel Nestra, EST2006-02 Mariusz Ska'ba, POL2006-04 Du"an !ukic, SRB2006-05 Hojoo Lee, KOR2006-06 Finbarr Holland, IRL – IMO32006-08 Du"an !ukic, SRB – IMO22006-09 Federico Ardila, COL2006-12 Federico Ardila, COL2006-13 Kei Irie, JPN2006-14 Hojoo Lee, KOR – IMO12006-15 Vyacheslav Yasinskiy, UKR2006-16 Zuming Feng, USA2006-18 Dimitris Kontogiannis, HEL2006-20 Tomá" Jurík, SVK2006-21 Waldemar Pompe, POL2006-23 Du"an !ukic, SRB – IMO62006-24 Zuming Feng, USA – IMO42006-25 J.P. Grossman, CAN2006-26 Johan Meyer, SAF2006-27 Dan Schwarz, ROU – IMO52006-29 Zoran %unic, USA2006-30 Juhan Aru, EST

2007-01 Michael Albert, NZL – IMO12007-02 Nikolai Nikolov, BGR2007-03 Juhan Aru, EST2007-05 Vjekoslav Kovac, HRV2007-06 Waldemar Pompe, POL

2007-07 G. Woeginger, NLD – IMO62007-08 Du"an !ukic, SRB2007-09 Kei Irie, JPN2007-10 Gerhard Woeginger, NLD2007-11 Omid Hatami, IRN2007-12 R. Gologan, D. Schwarz, ROU2007-13 Vasiliy Astakhov, RUS – IMO32007-14 Gerhard Woeginger, AUT2007-15 Vyacheslav Yasinskiy, UKR2007-16 Marek Pechal, CZE – IMO42007-17 Farzin Barekat, CAN2007-18 Vyacheslav Yasinskiy, UKR2007-19 Charles Leytem, LUX – IMO22007-20 Christopher Bradley, UNK2007-21 Z. Feng, O. Golberg, USA2007-22 Davoud Vakili, IRN2007-23 Waldemar Pompe, POL2007-24 Stephan Wagner, AUT2007-25 Dan Brown, CAN2007-26 Gerhard Woeginger, NLD2007-27 Jerzy Browkin, POL2007-28 M. Jamali,

N. Ahmadi Pour Anari, IRN2007-29 K. Buzzard,

E. Crane, UNK – IMO52007-30 N.V. Tejaswi, IND

2008-01 Hojoo Lee, KOR – IMO42008-02 Walther Janous, AUT – IMO22008-05 Pavel Novotn&, SVK2008-06 (ymantas Darbenas, LTU2008-09 Vidan Govedarica, SRB2008-10 Jorge Tipe, PER2008-11 B. Le Floch,

I. Smilga, FRA – IMO52008-14 A. Gavrilyuk, RUS – IMO12008-15 Charles Leytem, LUX2008-16 John Cuya, PER2008-19 Du"an !ukic, SRB2008-20 V. Shmarov, RUS – IMO62008-21 Angelo Di Pasquale, AUS2008-24 Du"an !ukic, SRB2008-26 K. Cesnavicius, LTU – IMO3

2009-01 Michal Rolínek, CZE2009-03 Bruno Le Floch, FRA – IMO5


2009-06 Gabriel Carroll, USA – IMO32009-07 Japanese PSC, JPN2009-08 Michael Albert, NZL2009-14 D. Khramtsov, RUS – IMO62009-15 Gerhard Woeginger, AUT2009-16 H. Lee,

P. Vandendriessche,J. Vonk, BEL – IMO4

2009-17 Sergei Berlov, RUS – IMO22009-19 David Monk, UNK2009-21 Eugene Bilopitov, UKR2009-24 Ross Atkins, AUS – IMO12009-25 Jorge Tipe, PER2009-28 József Pelikán, HUN2009-29 Okan Tekman, TUR

References

1. M. Aassila, 300 Défis Mathématiques, Ellipses, 2001.2. M. Aigner, G.M. Ziegler, Proofs from THE BOOK, Springer; 4th edition, 2009.3. G.L. Alexanderson, L.F. Klosinski, L.C. Larson, The William Lowell Putnam Mathemat-

ical Competition, Problems and Solutions: 1965–1984, The Mathematical Associationof America, 1985.

4. N. Altshiller-Court, College Geometry: An Introduction to the Modern Geometry of theTriangle and the Circle, Dover Publications, 2007.

5. T. Andreescu, D. Andrica, 360 Problems for Mathematical Contests, GIL PublishingHouse, Zalau, 2003.

6. T. Andreescu, D. Andrica, An Introduction to Diophantine Equations, GIL PublishingHouse, 2002.

7. T. Andreescu, D. Andrica, Complex Numbers from A to ... Z, Birkhäuser, Boston, 2005.8. T. Andreescu, D. Andrica, Z. Feng, 104 Number Theory Problems, Birkhauser, Boston,2006.

9. T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu, Old and New Inequalities, GILPublishing House, 2004.

10. T. Adnreescu, G. Dospinescu, Problems from the Book, XYZ Press, 2008.11. T. Andreescu, B. Enescu, Mathematical Treasures, Birkhäuser, Boston, 2003.12. T. Andreescu, Z. Feng, 102 Combinatorial Problems, Birkhäuser Boston, 2002.13. T. Andreescu, Z. Feng, 103 Trigonometry Problems: From the Training of the USA IMO

Team, Birkhäuser Boston, 2004.14. T. Andreescu, Z. Feng, Mathematical Olympiads 1998–1999, Problems and Solutions

from Around the World, The Mathematical Association of America, 2000.15. T. Andreescu, Z. Feng, Mathematical Olympiads 1999–2000, Problems and Solutions

from Around the World, The Mathematical Association of America, 2002.16. T. Andreescu, Z. Feng, Mathematical Olympiads 2000–2001, Problems and Solutions

from Around the World, The Mathematical Association of America, 2003.17. T. Andreescu, Z. Feng, A Path to Combinatorics for Undergraduates: Counting Strate-

gies, Birkhauser, Boston, 2003.18. T. Andreescu, R. Gelca,Mathematical Olympiad Challenges, Birkhäuser, Boston, 2000.19. T. Andreescu, K. Kedlaya, P. Zeitz,Mathematical Contests 1995-1996, Olympiads Prob-

lems and Solutions from Around the World, American Mathematics Competitions, 1997.20. T. Andreescu, K. Kedlaya,Mathematical Contests 1996-1997, Olympiads Problems and

Solutions from Around the World, American Mathematics Competitions, 1998.

806 References

21. T. Andreescu, K. Kedlaya,Mathematical Contests 1997-1998, Olympiads Problems andSolutions from Around the World, American Mathematics Competitions, 1999.

22. T. Andreescu, O. Mushkarov, L. Stoyanov, Geometric Problems on Maxima and Min-ima, Birkhauser Boston, 2005.

23. M. Arsenovic, V. Dragovic, Functional Equations (in Serbian), Mathematical Society ofSerbia, Beograd, 1999.

24. M. A"ic et al., International Mathematical Olympiads (in Serbian), Mathematical Soci-ety of Serbia, Beograd, 1986.

25. M. A"ic et al., 60 Problems for XIX IMO (in Serbian), Society of Mathematicians, Physi-cists, and Astronomers, Beograd, 1979.

26. A. Baker, A Concise Introduction to the Theory of Numbers, Cambridge UniversityPress, Cambridge, 1984.

27. E.J. Barbeau, Polynomials, Springer, 2003.28. E.J. Barbeau, M.S. Klamkin, W.O.J. Moser, Five Hundred Mathematical Challenges,

The Mathematical Association of America, 1995.29. E.J. Barbeau, Pell’s Equation, Springer-Verlag, 2003.30. M. Becheanu, International Mathematical Olympiads 1959–2000. Problems. Solutions.

Results, Academic Distribution Center, Freeland, USA, 2001.31. E.L. Berlekamp, J.H. Conway, R.K. Guy, Winning Ways for Your Mathematical Plays,

Volumes 1-4, AK Peters, Ltd., 2nd edition, 2001 – 2004.32. G. Berzsenyi, S.B. Maurer, The Contest Problem Book V, The Mathematical Association

of America, 1997.33. R. Brualdi, Introductory Combinatorics, 4th edition, Prentice-Hall, 2004.34. C.J. Bradley, Challenges in Geometry : for Mathematical Olympians Past and Present,

Oxford University Press, 2005.35. P.S. Bullen, D.S. Mitrinovic , M. Vasic,Means and Their Inequalities, Springer, 1989.36. C. Chuan-Chong, K. Khee-Meng, Principles and Techniques in Combinatorics, World

Scientific Publishing Company, 1992.37. H.S.M. Coxeter, Introduction to Geometry, John Willey & Sons, New York, 196938. H.S.M. Coxeter, S.L. Greitzer, Geometry Revisited, Random House, New York, 1967.39. I. Cuculescu, International Mathematical Olympiads for Students (in Romanian), Edi-

tura Tehnica, Bucharest, 1984.40. A. Engel, Problem Solving Strategies, Springer, 1999.41. D. Fomin, A. Kirichenko, Leningrad Mathematical Olympiads 1987–1991, MathPro

Press, 1994.42. A.A. Fomin, G.M. Kuznetsova, International Mathematical Olympiads (in Russian),

Drofa, Moskva, 1998.43. A. Gardiner, The Mathematical Olympiad Handook, Oxford, 1997.44. R. Gelca, T. Andreescu, Putnam and Beyond, Springer 2007.45. A.M. Gleason, R.E. Greenwood, L.M. Kelly, The William Lowell Putnam Mathematical

Competition, Problems and Solutions: 1938–1964, The Mathematical Association ofAmerica, 1980.

46. R.K. Guy, Unsolved Problems in Number Theory, Springer, 3rd edition, 2004.47. S.L. Greitzer, International Mathematical Olympiads 1959–1977, M.A.A., Washington,

D. C., 1978.48. R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, 2nd Edition, Addison-

Wesley, 1989.49. L-S. Hahn, Complex Numbers & Geometry, New York, 1960.50. L. Hahn, New Mexico Mathematics Contest Problem Book, University of New Mexico

Press, 2005.

References 807

51. G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, Cambridge University Press; 2ndedition, 1998.

52. G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, Oxford UniversityPress; 5th edition, 1980.

53. R. Honsberger, From Erdos to Kiev: Problems of Olympiad Caliber, MAA, 1996.54. R. Honsberger, In Polya’s Footsteps: Miscelaneous Problems and Essays, MAA, 1997.55. D. Hu, X. Tao, Mathematical Olympiad – the Problems Proposed to the 31st IMO (in

Chinese), Peking University Press, 1991.56. V. Jankovic, Z. Kadelburg, P. Mladenovic, International and Balkan Mathematical

Olympiads 1984–1995 (in Serbian), Mathematical Society of Serbia, Beograd, 1996.57. V. Jankovic, V. Micic, IX & XIX International Mathematical Olympiads, Mathematical

Society of Serbia, Beograd, 1997.58. R.A. Johnson, Advanced Euclidean Geometry, Dover Publications, 1960.59. Z. Kadelburg, D. Djukic, M. Lukic, I. Matic, Inequalities (in Serbian), Mathematical

Society of Serbia, Beograd, 2003.60. N.D. Kazarinoff,Geometric Inequalities, Mathematical Association of America (MAA),

1975.61. K.S. Kedlaya, B. Poonen, R. Vakil, The William Lowell Putnam Mathematical Compe-

tition 1985–2000 Problems, Solutions and Commentary, The Mathematical Associationof America, 2002.

62. A.P. Kiselev (author), A. Givental (editor), Kiselev’s Geometry / Book I Planimetry,Sumizdat, 2006.

63. A.P. Kiselev (author), A. Givental (editor), Kiselev’s Geometry / Book II Stereometry,Sumizdat, 2008.

64. M.S. Klamkin, International Mathematical Olympiads 1979–1985 and Forty Supple-mentary Problems, M.A.A., Washington, D.C., 1986.

65. M.S. Klamkin, International Mathematical Olympiads 1979–1986, M.A.A., Washing-ton, D.C., 1988.

66. M.S. Klamkin, USA Mathematical Olympiads 1972–1986, M.A.A., Washington, D.C.,1988.

67. M.E. Kuczma, 144 Problems of the Austrian-Polish Mathematics Competition 1978–1993, The Academic Distribution Center, Freeland, Maryland, 1994.

68. J. Kurshak, 1Hungarian Problem Book I, MAA, 1967.69. J. Kurshak, 1Hungarian Problem Book II, MAA, 1967.70. A. Kupetov, A. Rubanov, Problems in Geometry, MIR, Moscow, 1975.71. S. Lando, Lectures on Generating Functions, AMS, 2003.72. H.-H. Langmann, 30th International Mathematical Olympiad, Braunschweig 1989, Bil-

dung und Begabung e.V., Bonn 2, 1990.73. L.C. Larson, Problem Solving Through Problems, Springer, 1983.74. H. Lausch, C. Bosch-Giral Asian Pacific Mathematics Olympiads 1989–2000, AMT,

Canberra, 2000.75. H. Lausch, P. Taylor, Australian Mathematical Olympiads 1979–1995, AMT, Canberra

1997.76. P.K. Hung, Secrets in Inequalities, GIL Publishing House, 2007.77. A. Liu, Chinese Mathematical Competitions and Olympiads 1981–1993, AMT, Can-

berra 1998.78. A. Liu, Hungarian Problem Book III, MAA, 2001.79. E. Lozansky, C. Rousseau, Winning Solutions, Springer-Verlag, New York, 1996.80. V. Micic, Z. Kadelburg, D. Djukic, Introduction to Number Theory (in Serbian), 4th

edition, Mathematical Society of Serbia, Beograd, 2004.

808 References

81. D.S. Mitrinovic , J. Pecaric, A.M. Fink, Classical and New Inequalities in Analysis,Springer, 1992.

82. D.S. Mitrinovic, J.E. Pecaric, V. Volenec, Recent Advances in Geometric Inequalities,Kluwer Academic Publishers, 1989.

83. P. Mladenovic, Combinatorics, 3rd edition, Mathematical Society of Serbia, Beograd,2001.

84. P.S. Modenov, Problems in Geometry, MIR, Moscow, 1981.85. P.S. Modenov, A.S. Parhomenko, Geometric Transformations, Academic Press, New

York, 1965.86. L. Moisotte, 1850 exercices de mathémathique, Bordas, Paris, 1978.87. L.J. Mordell, Diophantine Equations, Academic Press, London and New York, 1969.88. E.A. Morozova, I.S. Petrakov, V.A. Skvortsov, International Mathematical Olympiads

(in Russian), Prosveshchenie, Moscow, 1976.89. I. Nagell, Introduction to Number Theoury, John Wiley & Sons, Inc., New York, Stock-

holm, 1951.90. I. Niven, H.S. Zuckerman, H.L. Montgomery, An Introduction to the Theory of Numbers,

John Wiley and Sons, Inc., 1991.91. G. Polya, How to Solve It: A New Aspect of Mathematical Method, Princeton University

Press, 2004.92. C.R. Pranesachar, Shailesh A. Shirali, B.J. Venkatachala, C.S. Yogananda,Mathematical

Challenges from Olympiads, Interline Publishing Pvt. Ltd., Bangalore, 1995.93. V.V. Prasolov, Problems of Plane Geometry, Volumes 1 and 2, Nauka, Moscow, 1986.94. V.V. Prasolov, V.M. Tikhomirov Geometry, Volumes 1 and 2, American Mathematical

Society, 2001.95. A.S. Posamentier, C.T. Salking, Challenging Problems in Algebra, Dover Books in

Mathematics, 1996.96. A.S. Posamentier, C.T. Salking, Challenging Problems in Geometry, Dover Books in

Mathematics, 1996.97. I. Reiman, J. Pataki, A. Stipsitz, International Mathematical Olympiad: 1959–1999, An-

them Press, London, 2002.98. I.F. Sharygin, Problems in Plane Geometry, Imported Pubn, 1988.99. W. Sierpinski, Elementary Theory of Numbers, Polski Academic Nauk, Warsaw, 1964.100. W. Sierpinski, 250 Problems in Elementary Number Theory, American Elsevier Publish-

ing Company, Inc., New York, PWN, Warsaw, 1970.101. A.M. Slinko, USSR Mathematical Olympiads 1989–1992, AMT, Canberra, 1998.102. C.G. Small, Functional Equations and How to Solve Them, Springer 2006.103. Z. Stankova, T. Rike, A Decade of the Berkeley Math Circle, American Mathematical

Society, 2008.104. R.P. Stanley, Enumerative Combinatorics, Volumes 1 and 2, Cambridge University

Press; New Ed edition, 2001.105. D. Stevanovic, M. Milo"evic, V. Baltic,Discrete Mathematics: Problem Book in Elemen-

tary Combinatorics and Graph Theory (in Serbian), Mathematical Society of Serbia,Beograd, 2004.

106. A.M. Storozhev, International Mathematical Tournament of the Towns, Book 5: 1997–2002, AMT Publishing, 2006.

107. P.J. Taylor, International Mathematical Tournament of the Towns, Book 1: 1980–1984,AMT Publishing, 1993.


References 809


110. P.J. Taylor, A.M. Storozhev, International Mathematical Tournament of the Towns, Book4: 1993–1997, AMT Publishing, 1998.

111. J. Tattersall, Elementary Number Theory in Nine Chapters, 2nd edition, Cambridge Uni-versity Press, 2005.

112. I. Tomescu, R.A. Melter, Problems in Combinatorics and Graph Theory, John Wiley &Sons, 1985.

113. I. Tomescu et al., Balkan Mathematical Olympiads 1984-1994 (in Romanian), Gil, Za-lau, 1996.

114. R. Vakil, A Mathematical Mosaic: Patterns and Problem Solving, 2nd edition, MAA,2007.

115. J.H. van Lint, R.M. Wilson, A Course in Combinatorics, second edition, CambridgeUniversity Press, 2001.

116. I.M. Vinogradov, Elements of Number Theory, Dover Publications, 2003.117. I.M. Vinogradov, The Method of Trigonometrical Sums in the Theory of Numbers, Dover

Books in Mathematics, 2004.118. H.S. Wilf, Generatingfunctionology, Academic Press, Inc.; 3rd edition, 2006.119. A.M. Yaglom, I.M. Yaglom, Challenging Mathematical Problems with Elementary So-

lutions, Dover Publications, 1987.120. I.M. Yaglom, Geometric Transformations, Vols. I, II, III, The Mathematical Association

of America (MAA), 1962, 1968, 1973.121. P. Zeitz, The Art and Craft of Problem Solving, Wiley; International Student edition,

2006.

1

Djuki! · Jankovi! M

ati! · Petrovi!


Du"an Djuki! · Vladimir Jankovi! · Ivan Mati! · Nikola Petrovi!

The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009, Second Edition


The IMO Compendium

Du"an Djuki!Vladimir Jankovi!Ivan Mati!Nikola Petrovi!

Mathematics

ABCDPBM

The IMO Com

pendium

A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009

Second Edition

! e IMO Compendium is the ultimate collection of challenging high school level mathemat-ics problems. It is an invaluable resource, not only for students preparing for competitions, but for anyone who loves and appreciates math. Training for mathematical olympiads is enjoyed by talented students throughout the world. Olympiads have become one of the primary ways to recognize and develop talented youth with a potential to excel in areas that require abstract thinking. Although the problems appearing at IMO do not involve advanced mathematics, they must be di! cult and their solutions must arise from creative and clever insights rather than tedious calculations.

In preparation for the distinguished International Mathematical Olympiad (IMO) competi-tion, each participating country selects the top six high school students every year through a series of national olympiads. " ese students are invited to participate in the IMO, usually held in July. " e IMO is a two-day contest where each day competitors are given three problems which they work on independently. " e IMO host country appoints a special committee to which each country submits up to six problems. From this composite “longlist” of prob-lems, a “shortlist” of #$-%& problems is created. " e jury, consisting of one professor from each country, makes the ' nal selection from the shortlist a few days before the IMO begins.

" e IMO has sparked a burst of creativity among enthusiasts to create new and interest-ing mathematics problems. It can be safely said that the IMO and shortlisted problems are among the well-cra( ed problems created in a given year. " is book attempts to gather all of these problems with their solutions. In addition, the book contains all the available longlist problems, for a total of more than #&&& problems.

From the reviews of the ' rst edition:

"" e International Mathematical Olympiad, or IMO is the premier international competi-tion for talented high school mathematics students. … " is book collects statements and solutions of all of the problems ever set in the IMO, together with many problems proposed for the contest. … serves as a vast repository of problems at the Olympiad level, useful both to students … and to faculty looking for hard elementary problems. No library will want to be without a copy, nor will anyone involved in mathematics competitions …"

— (Fernando Q. Gouvêa, MathDL, March, #&&))

2nd Ed.9 781441 998538

ISBN 978-1-4419-9853-8

Documents

1 The IMO Compendium - imomath.com · 1 Djuki ć · Jankovi ć Mati ć · Petrovi ć Problem Books in Mathematics Dušan Djukić · Vladimir Janković · Ivan Matić · Nikola Petrović